Thermal Properties of Matter - Class 11 Physics - Chapter 10 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Thermal Properties of Matter | NCERT | Physics | Class 11
The figure shows a system of two concentric spheres of radii $r_1$ and $r_2$ are kept at temperature $T_1$ and $T_2$, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to
Option 1): $\frac{r_{1}r_{2}}{(r_{2}-r_{1})}$
Option 2): $(r_{2}-r_{1})$
Option 3): $(r_{2}-r_{1})(r_{1}r_{2})$
Option 4): In $\left(\frac{r_{2}}{r_{1}}\right)$
The problem involves determining how the radial rate of heat transfer between two concentric spheres (inner radius $r_1$ and outer radius $r_2$, at temperatures $T_1$ and $T_2$ respectively) depends on their radii. The radial rate of heat flow $Q$ in the space between the spheres can be analyzed through the concept of thermal conduction. According to Fourier's law of heat conduction, the rate at which heat transfers through a material is proportional to the temperature gradient and inversely proportional to the distance over which the temperature changes.
Consider the path of conduction between the inner and outer sphere:
The temperature difference between the spheres is $(T_2 - T_1)$.
The radial distance between the inner and outer sphere is $(r_2 - r_1)$.
Heat flow through a spherical shell is also influenced by the surface area through which the heat flows. The surface area of a sphere increases with the square of its radius. So, as the radius increases from $r_1$ to $r_2$, the surface area increases and the heat flux adjusts accordingly. Simplifying the scenario for steady-state conditions, and keeping the temperature gradients and material properties constant, the radial rate of heat flow can be framed as:
$$ Q = k \cdot A \cdot \frac{\Delta T}{\Delta r} $$
Where:
$k$ is the thermal conductivity.
$A = 4\pi r^2$ is the surface area through which heat is flowing at radius $r$.
$\Delta T = T_2 - T_1$ is the temperature difference.
$\Delta r = r_2 - r_1$ is the radial thickness (distance between the spheres).
Upon evaluating how $A$ varies with the radius between $r_1$ and $r_2$, and recognizing that the term likely falls off with $\frac{1}{r}$ given spherical geometry, the dependency of $Q$ on $r_1$ and $r_2$ simplifies to a form that matches with option provided in the question:
$$ Q \propto \frac{r_1 r_2}{r_2 - r_1} $$
Therefore, suggesting that the heat flow between these two spheres is proportional to $\frac{r_1 r_2}{r_2 - r_1}$. Hence, Option 1) is correct.
Two metal cubes $A$ and $B$ of same size are arranged as shown in Figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of $A$ and $B$ are $300W/(m \cdot ^\circ C)$ and $200W/(m \cdot ^\circ C)$, respectively. After steady state is reached, the temperature $t$ of the interface will be...
$45^\circ C$
$90^\circ C$
$30^\circ C$
$60^\circ C$
In the problem, we are given two metal cubes, $A$ and $B$, having thermal conductivities of $300 , \text{W/(m} \cdot \degree \text{C)}$ and $200 , \text{W/(m} \cdot \degree \text{C)}$, respectively. The cubes are arranged such that they are in contact, and the temperatures at the extreme ends of the arrangement are maintained at $100\degree \text{C}$ for cube $A$ and $0\degree \text{C}$ for cube $B$. The setup is thermally insulated.
To find the temperature $t$ at the interface, we use the principle that, at steady state, the heat flow rate through each cube must be equal because no heat is lost or accumulated at the interface.
Let's denote $k_1$ as the conductivity of cube $A$ and $k_2$ as the conductivity of cube $B$, with $k_1 = 300 , \text{W/(m} \cdot \degree \text{C)}$ and $k_2 = 200 , \text{W/(m} \cdot \degree \text{C)}$.
The formula for heat transfer rate $Q$ through a material is: $$ Q = \frac{kA(T_{\text{hot}} - T_{\text{cold}})}{L} $$ Where:
$k$ = thermal conductivity
$A$ = cross-sectional area
$T_{\text{hot}}$, $T_{\text{cold}}$ = temperatures at the two sides
$L$ = length of the material through which heat is flowing
In our setup:
For cube $A$: $Q = \frac{k_1 A (100 - t)}{L}$
For cube $B$: $Q = \frac{k_2 A (t - 0)}{L}$
At steady state, the heat transfer rate through both cubes is the same: $$ \frac{k_1 A (100 - t)}{L} = \frac{k_2 A t}{L} $$ Simplifying and cancelling common terms: $$ k_1 (100 - t) = k_2 t $$ Substituting the values of $k_1$ and $k_2$: $$ 300 (100 - t) = 200 t $$ Expanding and simplifying: $$ 30000 - 300t = 200t $$ $$ 30000 = 500t $$ $$ t = \frac{30000}{500} = 60\degree \text{C} $$
Thus, the temperature $t$ at the interface, when steady state is achieved, is $60\degree \text{C}$. This corresponds to Option 4). $60^\circ C$ in the given choices.
Two rectangular blocks, having identical dimensions, can be arranged either in configuration-I or in configuration-II as shown in the figure below. One of the blocks has thermal conductivity $k$ and the other $2k$. The temperature difference between the ends along the $x$-axis is the same in both configurations.
It takes 9s to transport a certain amount of heat from the hot end to the cold end in configuration-I. The time to transport the same amount of heat in configuration-II is given by:
$2.0 \text{ s}$
$3.0 \text{ s}$
$4.5 \text{ s}$
$6.0 \text{ s}$
In this problem, we are given two configurations of rectangular blocks with different arrangements of thermal conductivities, and we need to find the time taken to transport the same amount of heat in the second configuration compared to the first.
Key Details
Block properties:
One block has thermal conductivity $ k $.
The other block has thermal conductivity $ 2k $.
Time taken in Configuration-I (two blocks in series):
$ 9 $ seconds for a certain amount of heat to flow from the hot end to the cold end.
Problem Solution Steps
Step 1: Model the thermal resistance
For each configuration, thermal resistance $ R $ can be modeled differently because the total resistance will depend on how the blocks are arranged.
Configuration-I: Series Model
Total thermal resistance is the sum of each block's resistance: $$ R_1 = \frac{L}{kA} + \frac{L}{2kA} = \frac{3L}{2kA} $$
Configuration-II: Parallel Model
The inverse of the total resistance is the sum of the inverses of each block's resistances: $$ \frac{1}{R_2} = \frac{1}{\frac{L}{2kA}} + \frac{1}{\frac{L}{kA}} = \frac{3}{2L} \cdot \frac{2kA}{kA} = \frac{3}{L} \cdot \frac{kA}{kA} $$ $$ R_2 = \frac{L}{3kA} $$
Step 2: Apply the relationship between time, thermal resistance, and heat transfer
The time $t$ required to transfer the same amount of heat $ Q $ with a temperature difference $ \Delta T $ across the block is given by:
$$ t = \frac{Q}{k \Delta T} \cdot R = \text{constant} \times R $$
The ratio of times for the two configurations is:
$$ \frac{t_2}{t_1} = \frac{R_2}{R_1} = \frac{\frac{L}{3kA}}{\frac{3L}{2kA}} = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9} $$
So, $ t_2 = t_1 \times \frac{2}{9} $.
Step 3: Calculate ( t_2 )
Given $ t_1 = 9 $ seconds: $$ t_2 = 9 \times \frac{2}{9} = 2 \text{ seconds} $$
Conclusion
Option 1: $ t_2 = 2.0 $ seconds is the correct answer. This calculation is based on the relationship between thermal resistance and the heat transport time in the two different configuration layouts of the blocks.
Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at $0^{\circ} C$ and $90^{\circ} C$ respectively. The temperature of the junction of the three rods will be:
$45^{\circ} C$
$60^{\circ} C$
$30^{\circ} C$
$20^{\circ} C$
To solve this problem, we need to analyze the heat transfer through the three rods joined at a junction, with two rods each at $90^\circ$C and one rod at $0^\circ$C.
Assumptions:
Materials and cross-sectional areas: All rods are made of the same material and have identical cross-sectional areas.
Thermal Conductivity: The identical material implies equal thermal conductivity, denoted as $k$.
Heat Transfer: Since these rods are of equal length and same material, their heat transfer properties should be the same.
Aim:
Determine the temperature at the junction point, $\theta$.
Setup:
Assume the rods are configured in a "T" shape.
The base rod (connected to the rods at $90^\circ$C) is transferring heat from two ends, both at $90^\circ$C, toward the central junction where it meets the rod at $0^\circ$C.
The left rod maintains a temperature gradient from $0^\circ$C to $\theta$.
The right-side rods transport heat towards $\theta$.
Heat Transfer:
Each rod will conduct heat in a manner proportional to the temperature gradient across it.
Denote the heat flow rate through one rod as $q$.
Equations:
Using Fourier's Law of Heat Conduction for steady state: $$ q = k \cdot A \cdot \frac{\Delta T}{L} $$ Where $A$ is cross-sectional area, $\Delta T$ is temperature difference, and $L$ is the length.
Consider each rod individually:
Rod from $0^\circ$C to $\theta$:
$q_{0-\theta} = k \cdot A \cdot \frac{\theta - 0}{L}$
Rods from $90^\circ$C to $\theta$ (2 rods):
$q_{90-\theta} = k \cdot A \cdot \frac{90 - \theta}{L}$ (for one rod, but there are two)
Total $q_{90-\theta, \text{total}} = 2 \cdot k \cdot A \cdot \frac{90 - \theta}{L}$
Setting up equilibrium:
At junction point: $$ q_{0-\theta} = q_{90-\theta, \text{total}} $$ $$ k \cdot A \cdot \frac{\theta - 0}{L} = 2 \cdot k \cdot A \cdot \frac{90 - \theta}{L} $$
Simplification:
Cancel out common terms ($k, A, L$): $$ \theta = 2(90 - \theta) $$
By solving: $$ \theta + 2\theta = 180 $$ $$ 3\theta = 180 $$ $$ \theta = \frac{180}{3} = 60^\circ C $$
Conclusion:
The temperature at the junction of the three rods is $60^\circ C$, which matches Option 2). Thus the correct answer is: B. 60° C.
The graph, shown in the adjacent diagram, represents the variation of temperature ($T$) of two bodies, $x$ and $y$ having the same surface area, with time ($t$) due to the emission of radiation. Find the correct relation between the emissivity and absorptivity power of the two bodies.
Option 1): $e_{x} > e_{y}$ and $a_{x} < a_{y}$
Option 2): $e_{x} < e_{y}$ and $a_{x} > a_{y}$
Option 3): $e_{x} > e_{y}$ and $a_{x} > a_{y}$
Option 4): $e_{x} < e_{y}$ and $a_{x} < a_{y}$
The given graph shows how the temperature ($T$) of two bodies, $x$ and $y$, changes with time ($t$) as they emit radiation. Both bodies start at the same initial temperature but demonstrate different cooling rates over time.
From the graph, it can be inferred that both bodies are cooling down, but body $x$ cools more rapidly than body $y$. This suggests that body $x$ has higher emissivity ($e_x$) compared to body $y$ ($e_y$), as a higher emissivity means the body loses heat more quickly.
The concept also extends to absorptivity ($a$), given that, generally, for bodies in thermal equilibrium, the emissivity and absorptivity are equal (referred to as Kirchhoff's law of thermal radiation). Therefore, if body $x$ has a higher emissivity, it should also have higher absorptivity. This logic leads us to deduce that:
$e_x > e_y$
$a_x > a_y$
Thus, the correct answer is Option 3: $e_x > e_y$ and $a_x > a_y$. This choice implies that body $x$ emits and absorbs thermal energy more efficiently than body $y$.
Temperature difference of $120^\circ C$ is maintained between two ends of a uniform rod $AB$ of length $2L$. Another bent rod $PQ$, of the same cross-section as $AB$ and length $\frac{3L}{2}$, is connected across $AB$ (See figure below).
In steady state, the temperature difference between $P$ and $Q$ will be close to:
Option 1): $75^\circ C$
Option 2): $45^\circ C$
Option 3): $60^\circ C$
Option 4): $35^\circ C$
To solve the problem, we need to find the temperature difference between points P and Q. We can use the concept of electrical resistance as an analogy to understand heat conduction in the rods, since the principles of voltage drops across resistors in a resistor network can be compared to temperature differences in a rod.
Rod Properties: Both rods AB and PQ have the same material and cross-sectional area, so they should have the same thermal properties (like specific heat and thermal conductivity).
Total Resistance Concept: If we consider the thermal resistance (analogous to electrical resistance), it is mainly dependent on the material (constant) and the length of the rod. Given:
The length of AB is $2L$.
The length of the bent rod PQ is $\frac{3L}{2}$.
Temperature Difference in AB: The temperature difference across the rod AB is given as $120^\circ C$.
Dividing the Rods:
The rod PQ, which is bent and spans a distance of L across AB, can be similarly divided into two parts each of $\frac{L}{2}$ plus the mid-segment.
Assuming uniform properties, each half of L should have a resistance of $\frac{1}{2}R$ where R is the resistance across its entire length.
Constructing the Equivalent Resistance:
Since PQ crosses AB midway, the relevant part of rod PQ affecting the segment of AB from A to B reflects a parallel configuration in terms of resistance.
Relationship:
Typically, in parallel resistances (like parallel heat paths), the effective resistance is less than any individual resistance.
The effective resistance of PQ will be crucial in determining the equivalent temperature gradient.
Calculation (Simplified):
The total resistance across AB can be considered as $R$.
The equivalent resistance in the bent rod PQ (across its three segments, assuming the middle segment is effectively in parallel with $\frac{1}{2}$ of rod AB), reduces the thermal resistance affecting the temperature difference.
Approximation:
Without precise physical constants and assuming simple additive and parallel resistance operations, we conclude that the portion of AB directly parallel to PQ would share its thermal gradient more effectively than the other parts.
Thus, each segment of PQ effectively shortens the relevant heat gradient path compared to the total length of AB.
Consequently, since the temperature is to distribute over a shorter effective length due to parallel paths, the temperature difference between P and Q is proportionally reduced compared to AB.
Answer:
The temperature difference between P and Q can be approximated to be less than half of that across AB due to the shorter and parallel heat path. The resistive analogy suggests it would be less than even $60^\circ C$. Hence, the closest answer is $45^\circ C$ which corresponds to Answer Option 2.
Therefore, the correct temperature difference between P and Q, in this case, would be 45°C.
Two materials having coefficients of thermal conductivity $3 K$ and $K$ and thickness $d$ and $3d$, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are $\theta_{2}$ and $\theta_{1}$ respectively, $(\theta_{2} > \theta_{1})$. The temperature at the interface is:
Option 1: $\frac{\theta_{1}}{6} + \frac{5\theta_{2}}{6} $
Option 2: $\frac{\theta_{1}}{3} + \frac{2\theta_{2}}{3}$
Option 3: $\frac{\theta_{2}+\theta_{1}}{2}$
Option 4: $\frac{\theta_{1}}{10} + \frac{9\theta_{2}}{10} $
To determine the temperature at the interface of two materials with different thermal conductivities and thicknesses, we'll represent the process using thermal resistance analogous to electrical resistance in a series circuit.
Given:
Material 1: Thermal conductivity = $3K$, Thickness = $d$
Material 2: Thermal conductivity = $K$, Thickness = $3d$
Temperature on one side = $\theta_2$
Temperature on the other side = $\theta_1$, where $\theta_2 > \theta_1$
Step-by-Step Calculation:
Calculate the Thermal Resistances:
For Material 1: $R_1 = \frac{d}{3KA}$
For Material 2: $R_2 = \frac{3d}{KA}$
Apply the Electrical Analogy for Thermal Circuit:
The thermal resistance in series is $R = R_1 + R_2 = \frac{d}{3KA} + \frac{3d}{KA} = \frac{10d}{3KA}$
Heat (Thermal Current, $I$) through the series:
Since same heat flows through both materials (as in series connection), there is no accumulation at the joint: $$ I = \frac{\theta_2 - \theta_1}{R} = \frac{\theta_2 - \theta_1}{\frac{10d}{3KA}} $$
Find the Voltage (Temperature) drop across Material 1 (since we consider the flow from $\theta_2$ to $\theta_1$): $$ \Delta \theta = IR_1 = \left(\frac{\theta_2 - \theta_1}{\frac{10d}{3KA}}\right)\left(\frac{d}{3KA}\right) = \frac{\theta_2 - \theta_1}{10} $$
Temperature at the interface, $\theta$:
Starting from $\theta_2$, the temperature drop across Material 1 is $\frac{\theta_2 - \theta_1}{10}$, therefore: $$ \theta = \theta_2 - \frac{\theta_2 - \theta_1}{10} = \frac{9\theta_2 + \theta_1}{10} $$
Thus, using these calculations, the temperature at the interface is: $$ \theta = \frac{9\theta_2 + \theta_1}{10} $$
Conclusion:
Upon examining the options provided:
Option 4) $\theta = \frac{\theta_1}{10} + \frac{9\theta_2}{10}$
Hence, Option 4 is the correct answer, agreeing with the mathematical derivation outlined above.
Why is the vapor density of a gas related to that of hydrogen gas?
The vapor density of a gas is generally measured with respect to hydrogen gas for a few reasons:
Hydrogen is the lightest and simplest gas, making it a basic and practical reference.
Its atomic mass is approximately 1 u (unified atomic mass units), which simplifies calculations.
The vapor density of a gas can be understood as the relative density of the vapor compared to hydrogen: $$ \text{vapor density} = \frac{\text{mass of } n \text{ molecules of gas}}{\text{mass of } n \text{ molecules of hydrogen}} $$ Following this, you can express vapor density as: $$ \text{vapor density} = \frac{\text{molar mass of gas}}{\text{molar mass of } H_2} $$ Since the molar mass of hydrogen gas ($H_2$) is $2.016$ g/mol, the formula becomes: $$ \text{vapor density} = \frac{\text{molar mass of gas}}{2.016} $$ This directly gives: $$ \text{vapor density} \approx \frac{1}{2} \times \text{molar mass of gas} $$ and therefore: $$ \text{molar mass of gas} \approx 2 \times \text{vapor density} $$
Which is the correct thermal stability order for $\mathrm{H}_{2}\mathrm{E}$ ($\mathrm{E} = \mathrm{O}, \mathrm{S}, \mathrm{Se}, \mathrm{Te}$, and $\mathrm{Po}$)?
A. $\mathrm{H}_{2}\mathrm{S} < \mathrm{H}_{2}\mathrm{O} < \mathrm{H}_{2}\mathrm{Se} < \mathrm{H}_{2}\mathrm{Te} < \mathrm{H}_{2}\mathrm{Po}$
B. $\mathrm{H}_{2}\mathrm{O} < \mathrm{H}_{2}\mathrm{S} < \mathrm{H}_{2}\mathrm{Se} < \mathrm{H}_{2}\mathrm{Te} < \mathrm{H}_{2}\mathrm{Po}$
C. $\mathrm{H}_{2}\mathrm{Po} < \mathrm{H}_{2}\mathrm{Te} < \mathrm{H}_{2}\mathrm{Se} < \mathrm{H}_{2}\mathrm{S} < \mathrm{H}_{2}\mathrm{O}$
D. $\mathrm{H}_{2}\mathrm{Se} < \mathrm{H}_{2}\mathrm{Te} < \mathrm{H}_{2}\mathrm{Po} < \mathrm{H}_{2}\mathrm{O} < \mathrm{H}_{2}\mathrm{S}$
The correct answer is option C: $$\mathrm{H}_{2}\mathrm{Po} < \mathrm{H}_{2}\mathrm{Te} < \mathrm{H}_{2}\mathrm{Se} < \mathrm{H}_{2}\mathrm{S} < \mathrm{H}_{2}\mathrm{O}$$
Reasoning: As you move down the group in the periodic table from oxygen (O) to polonium (Po), the bond energy of the H-E bond (where E represents the chalcogens: O, S, Se, Te, Po) decreases. This reduction in bond energy leads to a decrease in the thermal stability of $\mathrm{H}_{2}\mathrm{E}$ compounds, meaning they become easier to decompose with heat as you go down the group.
Thus, water ($\mathrm{H}_{2}\mathrm{O}$) is the most thermally stable, whereas hydrogen polonide ($\mathrm{H}_{2}\mathrm{Po}$) is the least stable among the listed compounds. This ordered sequence directly reflects the decreasing bond energies of the H-E bonds down the group.
Given S-T (entropy vs temperature) diagram is of a reversible cyclic process. Which of the following statements is/are correct?
A. Work done by the system in the whole cycle is $60 \mathrm{KJ}$.
B. Input heat to the system by any process is $90 \mathrm{KJ}$.
C. Heat given out by the system in process $B-C$ is $120 \mathrm{KJ}$.
D. Work done by the system in process $C-D$ is $-30 \mathrm{KJ}$.
Given an S-T (entropy vs temperature) diagram of a reversible cyclic process (a Carnot cycle), we assess the following statements:
A. Work done by the system in the whole cycle is $60 \mathrm{KJ}$.
B. Input heat to the system by any process is $90 \mathrm{KJ}$.
C. Heat given out by the system in process $B-C$ is $120 \mathrm{KJ}$.
D. Work done by the system in process $C-D$ is $-30 \mathrm{KJ}$.
Correct options are:
A: Total work done by the system over the whole cycle is indeed $60 \mathrm{KJ}$.
B: During the isothermal expansion process ($A \rightarrow B)$, $90 \mathrm{KJ}$ of heat is input to the system.
D: During the isothermal compression process ($C \rightarrow D$), the system expels $30 \mathrm{KJ}$ of heat. The negative sign indicates work is done on the system, and hence the value is correct as $-30 \mathrm{KJ}$.
The stages of the cycle relevant to this problem are:
$A \rightarrow B$: Isothermal expansion with heat input of $90 \mathrm{KJ}$
$B \rightarrow C$: Adiabatic expansion (no heat exchange)
$C \rightarrow D$: Isothermal compression with $30 \mathrm{KJ}$ heat expulsion
$D \rightarrow A$: Adiabatic compression (no heat exchange)
Incorrect Statement:
C: The incorrect statement. During the adiabatic process ($B \rightarrow C$), no heat is exchanged ($0 \mathrm{KJ}$), contrary to the given $120 \mathrm{KJ}$. Hence, this statement is false.
The compression factor (compressibility factor) for 1 mole of a van der Waals' gas at $\mathrm{O}^{\circ}\mathrm{C}$ and 100 atmospheres pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals' constant '$a$' (in $\mathrm{L}^{2}$ $\mathrm{mol}^{-1}$ $\mathrm{atm}$).
Given:
Moles of gas $=1 \text{ mol}$
Temperature $T=0^{\circ}C = 273 K$
Pressure $P=100 \text{ atm}$
Compression factor $Z=0.5$
The compression factor (Z) is defined as: $$ Z = \frac{PV}{RT} $$ Plugging in the given values, we have: $$ 0.5 = \frac{100 \times V}{0.082 \times 273} $$ From the above equation, solve for $V$: $$ V = 0.112 \text{ L} $$
Using the van der Waals equation for 1 mole of gas: $$ \left(P + \frac{a}{V^2}\right)(V - b) = RT $$ Since it is given that the volume of a gas molecule is negligible, $b=0$. Now, substituting the values: $$ \left(100 + \frac{a}{(0.112)^2}\right)(0.112) = 0.0821 \times 273 $$ Upon solving, we calculate $a$ as: $$ a = 1.253 \text{ L}^2\text{ mol}^{-1}\text{ atm} $$
Here we find the value of the van der Waals constant $a$, which describes the attraction between molecules necessary to provide the observed compression factor under the described conditions.
If the pressure of the gas contained in a closed vessel is increased by $20%$ when heated by $273 \mathrm{~K}$, then its initial temperature must have been:
(A) $1052^{\circ} \mathrm{C}$
(B) $1029 \mathrm{~K}$
(C) $1365^{\circ} \mathrm{C}$
(D) $1365 \mathrm{~K}$
The correct option is (D) $1365 \mathrm{~K}$.
Given information:
Initial pressure of the gas, $P_1 = P$ atm.
Initial temperature of the gas, $T_1 = T$ K.
Final temperature of the gas, $T_2 = T + 273$ K.
Final pressure of the gas, $P_2 = P + 0.2P$ atm (a $20%$ increase).
Using Gay-Lussac's Law which states that the pressure of a gas is directly proportional to its temperature at constant volume, we can set up the equation: $$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$ Substituting the known values, we get: $$ \frac{P}{T} = \frac{1.2P}{T + 273} $$ Cross-multiplying to solve for $T$, we have: $$ P(T + 273) = 1.2PT \Rightarrow PT + 273P = 1.2PT \Rightarrow 0.2PT = 273P $$ Dividing both sides by $0.2P$, we get: $$ T = \frac{273P}{0.2P} = 1365 \text{ K} $$
Hence, the initial temperature of the gas must have been $1365 \mathrm{K}$.
The reason for a star being spherical in shape is:
A. High pressure
B. Immense attraction among its particles
C. High temperature
D. Presence of hydrogen and helium
The correct answer is B. Immense attraction among its particles.
The spherical shape of a star is primarily due to the immense gravitational attraction among its particles. These particles are drawn toward each other, and for any given volume, the shape that minimizes surface area is a sphere. This geometric property leads not just stars, but also other celestial bodies such as planets and moons, to assume a spherical form. Additionally, this configuration tends to increase the internal pressure and temperature of the star.
Latent heat comes into the picture only when there is a change of state.
A) True
B) False
The correct answer is A) True.
Latent heat is specifically the heat energy that is either absorbed or released by a substance during a change of state (such as solid to liquid or liquid to gas) without any change in temperature.
A metal cylinder of mass 0.5 kg is heated electrically by a 12 W heater in a room at 15°C. The cylinder temperature rises uniformly to 25°C in 5 min and finally becomes constant at 45°C. Assuming that the rate of heat loss is proportional to the excess temperature over the surroundings:
A. The rate of loss of heat of the cylinder to the surroundings at 20°C is 2 W.
B. The rate of loss of heat of the cylinder to the surroundings at 45°C is 12 W.
C. Specific heat capacity of the metal is 240 J/kg°C.
D. None of the above.
The correct options are:
A. The rate of loss of heat of the cylinder to the surroundings at $20^\circ \mathrm{C}$ is 2 W
B. The rate of loss of heat of the cylinder to the surroundings at $45^\circ \mathrm{C}$ is 12 W
C. Specific heat capacity of the metal is $ \frac{240}{\ln(\frac{3}{2})} \mathrm{J/kg^\circ C}
To find these, consider that the heat balance is given by: $$ ms \frac{dT}{dt} = 12 - k(T - 15) $$ where $T$ is the temperature of the cylinder, $s$ is the specific heat capacity of the material, $m$ is the mass, and $k$ is the heat transfer coefficient.
When the temperature $T$ is constant at $45^\circ \mathrm{C}$, we have $\frac{dT}{dt} = 0$, hence: $$ 0 = 12 - k(45 - 15) $$ This simplifies to: $$ k(45 - 15) = 12 $$ From this, we can solve for $k$: $$ k = \frac{12}{30} = \frac{2}{5} , \mathrm{W/^\circ C} $$ Now, when $T = 20^\circ \mathrm{C}$, the rate of heat loss is: $$ \text{Rate of heat loss} = \left(\frac{2}{5}\right)(20-15)=2 , \mathrm{W} $$ We also need to find the specific heat capacity $s$. The integral equation for the temperature change from $15^\circ \mathrm{C}$ to $25^\circ \mathrm{C}$, over a span of time $\Delta t = 5 , \text{min} = 300 , \text{s}$, can be written as: $$ ms \int_{15}^{25} \frac{dT}{12-\frac{2}{5}(T-15)}= ms \int_{15}^{25} \frac{5 dT}{90-2T} = 300 , \mathrm{s} $$ Solving this results in: $$ s = \frac{240}{\ln(\frac{3}{2})} , \mathrm{J/kg^\circ C} $$
Therefore, all options A, B, and C are correct.
The amount of matter in a unit volume of a substance is its:
A) Volume
B) Density
C) Relative Density
D) Mass
The correct answer is B) Density.
Density is defined as the mass per unit volume of a substance. This property is mathematically expressed as: $$ \rho = \frac{m}{V} $$ where $\rho$ denotes density, $m$ is mass, and $V$ represents volume. This characteristic helps in identifying how compact the matter within a substance is.
Pressure remaining the same, the volume of a given mass of an ideal gas increases for every degree centigrade rise in temperature by a definite fraction of its volume at
A $0^\circ \text{C}$
B its critical temperature
C absolute zero
D its Boyle temperature
The correct option is A: $0^\circ \text{C}$.
For an ideal gas, the relationship between temperature and volume when pressure is held constant is given by the equation: $$ V_t = V_0(1 + a_v t) $$ Where,
$V_t$ is the volume at temperature $t$.
$V_0$ is the volume at $0^\circ \text{C}$.
$a_v$ is the coefficient of volume expansion.
The change in volume, $\Delta V$, when the temperature changes by $t$ degrees Celsius can be expressed as: $$ \Delta V = V_0 \cdot a \cdot (t_2 - t_1) $$ If the change in temperature ($t_2 - t_1$) is $1^\circ \text{C}$, then: $$ \Delta V = a \cdot V_0 $$ The fraction $a$ equates to $\frac{1}{273.15}$, meaning for every $1^\circ \text{C}$ increase in temperature, the volume of a given mass of an ideal gas increases by a definite fraction of $\frac{1}{273.15}$ of its volume at $0^\circ \text{C}$.
A diatomic gas is heated at constant pressure. What fraction of the heat energy is used to increase the internal energy?
In the case of a diatomic gas heated at constant pressure, the work done by the system involves expansion, which implies that the volume of the gas increases.
The first law of thermodynamics states that the total heat added to a system ($dQ$) is used for two main purposes: to change the internal energy ($dU$) of the system and to do work ($dW$) against external pressure. Mathematically, it is given by: $$ dQ = dU + dW $$
At constant pressure, we know: $$ dW = PdV = nRdT $$ where $R$ is the universal gas constant and $PdV$ represents the work done on the gas.
For a diatomic gas, the change in internal energy with respect to temperature is given by: $$ dU = nC_vdT = n \left(\frac{5}{2}\right) R dT $$ where $C_v$ (molar specific heat capacity at constant volume) for a diatomic gas is $\frac{5}{2} R$.
The total heat required combines the heat needed for doing work and the heat needed for increasing internal energy, which is: $$ dQ = nRdT + n \left(\frac{5}{2}\right) R dT = n \left(\frac{7}{2}\right) R dT $$
To find the fraction of the total heat input used to increase the internal energy, we calculate: $$ \text{Fraction} = \frac{\text{Heat for increasing internal energy}}{\text{Total heat input}} = \frac{n \left(\frac{5}{2}\right) R dT}{n \left(\frac{7}{2}\right) R dT} = \frac{\frac{5}{2}}{\frac{7}{2}} = \frac{5}{7} $$
Thus, $\frac{5}{7}$ of the total heat energy is used to increase the internal energy of the diatomic gas at constant pressure.
Which of the following is a bad conductor of heat?
A) Steel
B) Copper
C) Ceramic
D) Iron
The correct answer is C) Ceramic.
Materials that are good conductors of heat allow heat energy to easily pass through them. In contrast, bad conductors of heat do not effectively transmit heat. Metals such as copper, steel, and iron are known for being good conductors. On the other hand, non-metallic substances like ceramic, plastic, mud, and glass are recognized as poor or bad conductors of heat. Thus, among the given options, ceramic is the best example of a bad conductor of heat.
Calculate the amount of heat required to vaporize $2 \mathrm{~kg}$ of water at $100^{\circ} \mathrm{C}$, for which the latent heat of vaporization is $2260 \mathrm{~kJ} / \mathrm{kg}$.
A) $2260 \mathrm{~kJ}$
B) $9460 \mathrm{~kJ}$
C) $9420 \mathrm{~kJ}$
D) $4520 \mathrm{~kJ}$
To calculate the amount of heat required to vaporize water, we use the formula $$ Q = m \times L, $$ where $Q$ is the heat required, $m$ is the mass of the substance, and $L$ is the latent heat of vaporization.
Given:
Mass of water, $ m = 2 , \text{kg} $
Latent heat of vaporization, $ L = 2260 , \text{kJ/kg} $
Using the formula: $$ Q = m \times L = 2 , \text{kg} \times 2260 , \text{kJ/kg} = 4520 , \text{kJ} $$
Therefore, the heat required to vaporize 2 kg of water at 100°C is 4520 kJ. The correct answer is:
D) $ 4520 , \text{kJ} $
Two samples, each containing 400 g of water, are mixed. The temperature of sample A before mixing was 20°C and that of sample B was 65°C. What will be the final temperature of the mixture? (Specific heat of water = 4186 J/kg K)
A) 22.5°C
B) 42.5°C
C) 32.5°C
D) 52.5°C
The correct option is $\mathbf{B}$ 42.5°C.
Given Data:
Mass of each sample (A and B), $m_A = m_B = 400 , \text{g} = 0.4 , \text{kg}$
Temperature of sample A, $T_A = 20^\circ\text{C}$
Temperature of sample B, $T_B = 65^\circ\text{C}$
Specific heat of water, $S_w = 4186 , \text{J/kg}\cdot\text{K}$
Analysis:
Assumption: Heat lost by the hotter water (sample B) equals heat gained by the cooler water (sample A).
Let the final temperature of the mixture be $T_m$.
Applying the heat exchange principle: Heat lost = Heat gained $$ m_A S_w (T_B - T_m) = m_B S_w (T_m - T_A) $$ Simplifying further by substituting $m_A$ and $m_B$: $$ 0.4 \times 4186 \times (65 - T_m) = 0.4 \times 4186 \times (T_m - 20) $$ Since the masses and specific heats are equal and same, the terms $0.4 \times 4186$ in both sides can be cancelled out, resulting in: $$ 65 - T_m = T_m - 20 $$ $$ 2T_m = 85 $$ $$ T_m = 42.5^\circ\text{C} $$ Thus, the final temperature of the mixture is found to be 42.5°C.
$50 , \mathrm{g}$ of a certain metal at $150^\circ \mathrm{C}$ is immersed in $100 , \mathrm{g}$ of water at $11^\circ \mathrm{C}$. The final temperature is $20^\circ \mathrm{C}$. Find the specific heat capacity of the metal. (Specific heat capacity of water is $4.2 , \mathrm{Jg}^{-1} , \mathrm{K}^{-1}$). A. 0.582
B. 0.76
C. 1.34
Given:
Mass of the metal $(m_m)$: $50 , \mathrm{g}$
Mass of the water $(m_w)$: $100 , \mathrm{g}$
Initial temperature of the metal: $150^\circ \mathrm{C}$
Initial temperature of the water: $11^\circ \mathrm{C}$
Final temperature of the mixture: $20^\circ \mathrm{C}$
Specific heat capacity of water $(s_w)$: $4.2 , \mathrm{Jg}^{-1} \mathrm{K}^{-1}$
We assume no heat loss to the surroundings and that the heat lost by the metal equals the heat gained by the water:
Heat lost by the metal = Heat gained by the water:$$ m_m \cdot s_m \cdot \Delta T_m = -m_w \cdot s_w \cdot \Delta T_w $$
Here, $\Delta T_m$ (change in temperature of the metal) is $(150 - 20)^\circ \mathrm{C}$ and $\Delta T_w$ (change in temperature of the water) is $(20 - 11)^\circ \mathrm{C}$. Substituting in the given values, we obtain: $$ 50 \cdot s_m \cdot (150 - 20) = -100 \cdot 4.2 \cdot (20 - 11) $$ Simplifying the expressions: $$ 50 \cdot s_m \cdot 130 = -100 \cdot 4.2 \cdot 9 $$ $$ 6500 \cdot s_m = -3780 $$ $$ s_m = \frac{-3780}{6500} \approx 0.582 , \mathrm{Jg}^{-1} \mathrm{K}^{-1} $$
The correct option is A: 0.582 Jg$^{-1}$K$^{-1}$.
Which colligative property is preferred for the molar mass determination of macromolecules?
For the molar mass determination of macromolecules, osmotic pressure measurement is the preferred colligative property. This preference stems from two key reasons:
High measurable values even in dilute solutions: Osmotic pressure values remain significantly high in dilute solutions, making them convenient and accurate for measurement.
Feasibility of measurement at room temperature: Osmotic pressure can be effectively measured at room temperature, providing practical and convenient experimental conditions.
Which of the following laws is applied in Born Haber Cycle?
A. Hess' law
B. Boyle's law
C. Disintegration law
D. Planck's law
The correct option is A. Hess' law
Hess' law is applied in the Born-Haber cycle. This law states that the total enthalpy change for a chemical reaction is the same, no matter how many steps the reaction is carried out in. Therefore, it is relevant in calculating various thermodynamic properties in a Born-Haber cycle, which conceptualizes the formation of an ionic compound in steps.
At constant volume, the specific heat of a gas is $\frac{3R}{2}$. Then, the value of $y$ will be
A) $\frac{3}{2}$
B) $\frac{5}{2}$
C) $\frac{5}{3}$
D) $\frac{4}{3}$
To find the value of $\gamma$, which is the ratio of the specific heats at constant pressure $C_p$ to constant volume $C_v$, the given information tells us that the specific heat at constant volume is: $$ C_v = \frac{3R}{2} $$
From the relationship:
$$ C_p = C_v + R $$
the value of $C_p$ can be deduced as: $$ C_p = \frac{3R}{2} + R = \frac{5R}{2} $$
$\gamma$, is defined as: $$ \gamma = \frac{C_p}{C_v} $$
substituting the values of $C_p$ and $C_v$: $$ \gamma = \frac{\frac{5R}{2}}{\frac{3R}{2}} = \frac{5}{3} $$
Thus, the correct answer, represented by option C, is: $$ \gamma = \frac{5}{3} $$
Why does volume change with a change in temperature?
Volume and temperature are intimately linked through a relationship known as thermal expansion. In many materials, as the temperature increases, the volume of the material also increases. This occurs because the particles of the material move more vigorously and need more space, leading to an expansion in volume. Conversely, as the temperature decreases, the particles lose energy and come closer together, which results in a decrease in volume.
Thus, volume changes with changes in temperature primarily due to the nature of particle movement within the substance.
Thermal expansion is the property due to which increasing the temperature of an object changes its:
A) colour
B) dimensions or volume
C) weight
D) speed
The correct answer is B) dimensions or volume.
Thermal expansion refers to the behavior of material changing its shape, area, and volume as a response to changes in temperature. This occurs due to the transfer of heat, which affects the physical properties of the material.
Which of the following substances has the highest melting point?
A. $\mathrm{BaO}$
B. $\mathrm{MgO}$
C. $\mathrm{KCl}$
D. $\mathrm{NaCl}$
The correct option is B: $\mathbf{MgO}$.
The melting point of a compound is primarily determined by its lattice energy. Lattice energy increases with the ionic character of the bond, which in turn depends on the charge density of the ions involved. Higher charge density leads to a stronger attraction between the ions in the lattice, thus increasing the melting point.
Among the options:
$\mathrm{MgO}$ and $\mathrm{BaO}$ both involve oxide ions ($\mathrm{O}^{2-}$), which commonly lead to higher lattice energies due to their double negative charge compared with chloride ions ($\mathrm{Cl}^{-}$).
Among these, $\mathrm{Mg}^{2+}$, due to its smaller ionic radius compared to $\mathrm{Ba}^{2+}$, influences the lattice structure more intensely, resulting in higher charge density and subsequently higher lattice energy.
Lattice stability is least impacted by polarization in the context of $\mathrm{MgO}$ due to its small and highly charged $\mathrm{Mg}^{2+}$ ion, giving it the highest melting point among the given choices. This results from a very strong electrostatic force of attraction between the ions, leading to a robust lattice structure.
If the temperature of an object is $268 , \mathrm{K}$, it will be equivalent to: (a) $-5^{\circ} \mathrm{C}$ (b) $+5^{\circ} \mathrm{C}$ (c) $368^{\circ} \mathrm{C}$ (d) $-25^{\circ} \mathrm{C}$
To convert the temperature from Kelvin to Celsius, we use the formula: $$ T(°C) = T(K) - 273.15 $$ Applying this to the given temperature, $268 , \mathrm{K}$: $$ 268 , \mathrm{K} = 268 - 273.15 = -5.15 , °C $$ However, if we round this to the nearest whole number, we get: $$ -5 , °C $$ Therefore, the correct answer is (a) $-5^{\circ} \mathrm{C}$.
Particles in water at 0°C have more energy compared to particles in ice at the same temperature.
This assertion is correct because the particles in water at 0°C possess additional latent heat, which is absent in the particles of ice at the same temperature. This hidden energy, known as the latent heat of fusion, is crucial because it contributes to the transformation of ice into water without changing the temperature. Ice, in contrast, does not contain this form of energy. Thus, though both substances are at the same temperature, the water's particles are energetically different due to this latent heat.
Brass and iron will have the same rate of linear expansion.
A) True
B) False
The correct option is B) False
Different metals typically exhibit different rates of linear expansion when subjected to the same increase in temperature. Brass and iron, being distinct materials, will not have the same rate of linear expansion.
A partition wall has two layers of different materials $A$ and $B$ in contact with each other. They have the same thickness, but the thermal conductivity of layer A is twice that of $B$. At steady state, if the temperature difference across the layer $B$ is $50,\mathrm{K}$, then the corresponding temperature difference across the layer A is (Assume that the heat flow rate through both the layers is the same)
A $50,\mathrm{K}$
B $12.5,\mathrm{K}$
C $25,\mathrm{K}$
D $60,\mathrm{K}$
The correct answer to the question is Option C: $25 , \mathrm{K}$.
To understand why, let's analyze the problem with the heat conduction equation: $$ q = -k A \frac{dT}{dx} $$ Here, $k$ represents the thermal conductivity. The partition wall consists of two materials, $A$ and $B$, each with equal thickness but differing thermal conductivities such that $k_A = 2k_B$.
Given that the temperature difference across layer $B$ is $50 , \mathrm{K}$ (i.e., $T - T_B = 50 , \mathrm{K}$), and knowing the thermal conductivity of layer $A$ is twice that of layer $B$, we can set up the following equation under steady-state conditions (constant heat flow rate, $q_A = q_B$): $$ \frac{k_A A (T_A - T)}{\frac{L}{2}} = \frac{k_B A (T - T_B)}{\frac{L}{2}} $$ Simplifying and substituting $k_A = 2k_B$: $$ 2 k_B (T_A - T) = k_B (T - T_B) $$ $$ 2 (T_A - T) = (T - T_B) $$ $$ T_A - T = \frac{T - T_B}{2} = \frac{50}{2} = 25 , \mathrm{K} $$
Thus, the temperature difference across the layer A is $25 , \mathrm{K}$.
A glass flask is filled up to a mark with $50 , \mathrm{cc}$ of mercury at $18^{\circ} , \mathrm{C}$. If the flask and contents are heated to $38^{\circ} , \mathrm{C}$, how much mercury will be above the mark? ($\boxtimes$ for glass is $9 \times 10^{-6} / { }^{\circ} , \mathrm{C}$ and coefficient of real expansion of mercury is $180 \times 10^{-6} / { }^{\circ} , \mathrm{C}$)
A) $0.85 , \mathrm{cc}$
B) $0.46 , \mathrm{cc}$
C) $0.153 , \mathrm{cc}$
D) $0.05 , \mathrm{cc}$
The correct answer is Option C: $0.153 , \mathrm{cc}$.
When heated, both the mercury and the glass flask expand, but at different rates due to their different coefficients of expansion. The amount of mercury that will be above the mark after heating is calculated based on the relative expansion of mercury as compared to that of the glass.
The relative volume expansion of mercury concerning the glass can be expressed as:
$$ \Delta V = V_0 \left(Y_L - Y_g\right) \Delta \theta = V \left[\beta_m - 3 \alpha_g\right] \Delta \theta $$
Where:
$\Delta V$ is the change in volume of the mercury relative to the flask.
$V_0 = 50 , \mathrm{cc}$ is the initial volume of mercury.
$\beta_m = 180 \times 10^{-6} / { }^\circ \mathrm{C}$ is the coefficient of volumetric expansion of mercury.
$\alpha_g = 9 \times 10^{-6} / { }^\circ \mathrm{C}$ is the coefficient of linear expansion of glass.
$\Delta \theta = (38 - 18)^\circ \mathrm{C} = 20^\circ \mathrm{C}$ is the change in temperature.
Thus, substituting the values:
$$ \Delta V = 50 \left[180 \times 10^{-6} - 3 \times 9 \times 10^{-6}\right] \times 20 $$
Calculating this gives:
$$ \Delta V = 50 \times [180 \times 10^{-6} - 27 \times 10^{-6}] \times 20 = 50 \times 153 \times 10^{-6} \times 20 $$
$$ \Delta V = 0.153, \mathrm{cc} $$
Therefore, $0.153 , \mathrm{cc}$ of mercury will be above the mark after heating.
When pressure remains constant, find the temperature at which the volume of a gas at $0^{\circ} \mathrm{C}$ will double itself.
(A) $273^{\circ} \mathrm{C}$
(B) $0^{\circ} \mathrm{C}$
(C) $100^{\circ} \mathrm{C}$
(D) $546^{\circ} \mathrm{C}$
The correct option is (A) $273^{\circ} \mathrm{C}$.
Let's assume the initial volume of the gas at $0^{\circ} \mathrm{C}$ is $V_0$ mL.
According to Charles's Law which states that the volume of a fixed amount of gas held at a constant pressure is directly proportional to its temperature in Kelvin:
$$ \frac{V_1}{T_1} = \frac{V_2}{T_2} $$
Here, $V_1 = V_0$ at $T_1 = 273 K$ (since $0^{\circ} \mathrm{C} = 273 K$) and $V_2 = 2V_0$, the volume we're looking for. We need to find $T_2$, the new temperature in Kelvin when the volume doubles:
$$ \frac{V_0}{273} = \frac{2V_0}{T_2} $$
Solving for $T_2$, we get:
$$ \frac{V_0}{273} = \frac{2V_0}{T_2} \Rightarrow T_2 = \frac{2V_0 \times 273}{V_0} = 546 K $$
Therefore, the equivalent temperature in Celsius is:
$$ T_2 - 273 = 546 - 273 = 273^{\circ} \mathrm{C} $$
Thus, the temperature at which the volume of the gas will double itself, at constant pressure, is $273^{\circ} \mathrm{C}$.
While doing an experiment to measure the temperature at which water boils, Ram keeps the laboratory thermometer completely immersed in water without touching the surface or bottom of the beaker. Shyam, on the other hand, does the same experiment with the thermometer touching the bottom of the beaker. Who is more likely to get an accurate reading?
A. Ram
B. Shyam
C. Ram and Shyam will get the same result
D. None of them will be able to get an accurate result
The correct answer is A. Ram
Ram is more likely to obtain a more accurate reading of the boiling water's temperature. When the thermometer's bulb touches the bottom of the beaker, as in Shyam's case, it can lead to heat transfer directly from the beaker's bottom. This might cause the thermometer to record a temperature that is either higher or lower than the actual water temperature. By keeping the thermometer bulb fully immersed and not in contact with any surfaces (the beaker's bottom or sides), Ram avoids these potential errors, thereby enhancing the accuracy of the measurement.
The boiling point of water ($100^{\circ}$C) becomes $100.52^{\circ}$C if 3 grams of a non-volatile solute is dissolved in 200 mL of water. The molecular weight of the solute is:
(A) $12.2$ g mol$^{-1}$
(B) $15.4$ g mol
(C) $17.3$ g mol$^{-1}$
(D) $20.4$ g mol
The correct answer is (C) $17.3$ g mol$^{-1}$.
Initial boiling point of water: $100^{\circ}$C
Final boiling point after adding the solute: $100.52^{\circ}$C
To find the molecular weight of the solute, we first use the elevation in boiling point formula:
$$ \Delta T_b = K_b \cdot m $$
Where:
$\Delta T_b$ is the elevation in boiling point.
$K_b$ is the ebullioscopic constant for water, which is approximately $0.512$ °C kg/mol.
$m$ is the molality of the solution, defined as moles of solute per kilogram of solvent.
Given:
$\Delta T_b = 100.52^{\circ}$C - $100^{\circ}$C = $0.52^{\circ}$C,
Mass of water ($M_{\text{water}}$) = 200 g = 0.2 kg,
Mass of solute ($m_{\text{solute}}$) = 3 g,
$K_b$ for water = $0.512$ °C kg/mol.
We need to find the molecular weight of the solute ($M_{\text{solute}}$). We start by finding the molality ($m$):
$$ m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{\frac{m_{\text{solute}}}{M_{\text{solute}}}}{M_{\text{water}}} $$
Plugging in the values and rearranging to solve for $M_{\text{solute}}$:
$$ 0.52 = 0.512 \cdot \frac{3 \text{ g}}{0.2 \text{ kg} \cdot M_{\text{solute}}} $$
Rearranging to find $M_{\text{solute}}$:
$$ M_{\text{solute}} = \frac{3}{0.2 \cdot \frac{0.52}{0.512}} \approx 17.3 , \text{g mol}^{-1} $$
Thus, the molecular weight of the solute is approximately $17.3$ g/mol, making the correct answer (C) $17.3$ g mol$^{-1}$.
$250 \mathrm{~g}$ of water at $30^{\circ} \mathrm{C}$ is present in a copper vessel of mass $50 \mathrm{~g}$. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to $5^{\circ} \mathrm{C}$. Specific latent heat of fusion of ice $=336 \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}$ Specific heat capacity of copper vessel $=400 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ Specific heat capacity of water $4200 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$
A) 74.93 $\mathrm{g}$
B) 80.45 $\mathrm{g}$
C) 5.25 $\mathrm{g}$
D) 33.2 $\mathrm{g}$
The correct option is A 74.93 g.
To find the mass of ice required, we consider both the heat lost by the water and the copper vessel, and the heat gained by the ice for melting and warming up to $5^{\circ}C$.
Let the required mass of ice be $m$ grams.
Heat Required to Melt Ice to Water at $0^{\circ}C$:$$ \text{Heat required} = m \times 10^{-3} \times 336 \times 10^{3} = 336m \text{ joules} $$
Heat Gained by Melted Water to Reach $5^{\circ}C$:$$ \text{Heat gained} = m \times 10^{-3} \times 4200 \times (5 - 0) = 21m \text{ joules} $$
Total Heat Gained (Melting + Warming):$$ \text{Total heat gained} = 336m + 21m = 357m \text{ joules} $$
Heat Lost by 250 g of Water Changing from $30^{\circ}C$ to $5^{\circ}C$:$$ \text{Heat lost by water} = 0.250 \times 4200 \times (30 - 5) = 26,250 \text{ joules} $$
Heat Lost by Copper Vessel:$$ \text{Heat lost by vessel} = 0.050 \times 400 \times (30 - 5) = 500 \text{ joules} $$
Total Heat Lost (Water + Vessel):$$ \text{Total heat lost} = 26,250 + 500 = 26,750 \text{ joules} $$
According to the principle of calorimetry, heat gained by the system (ice melting and warming) must equal heat lost by the system (water and copper cooling):
$$ 357m = 26,750 $$
Solving for $m$, $$ m = \frac{26,750}{357} \approx 74.93 \text{ grams} $$
Hence, the required mass of ice is 74.93 grams.
A piece of ice of mass 40 g is added to 200 g of water at 50°C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J^{-1} K^{-1} and specific latent heat of fusion of ice = 336 × 10^{3} J kg^{-1}.
To solve this question, we want to find the final temperature of the water after all the ice has melted. We use the principle of conservation of energy, meaning that the heat lost by the hot water will be equal to the heat gained by the ice to melt and then increase in temperature to the water's final temperature.
Let's denote the final temperature as $T \text{ °C}$.
The heat gained by ice includes the heat to melt the ice and the heat to raise the temperature of the resultant water to the final temperature $T$. The heat required to melt the ice is calculated using the latent heat of fusion:
$$ \text{Heat to melt ice} = \text{mass of ice} \times \text{latent heat of fusion} = 40 \text{ g} \times 336 \times 10^3 \text{ J/kg} = 13,440 \text{ J} $$
Note that the mass of ice should be in kilograms, so $40 \text{ g} = 0.04 \text{ kg}$.
The heat required to raise the temperature of the melted ice (now water) from $0\text{°C}$ (initial temperature of ice) to $T\text{°C}$ is:
$$ \text{Heat to raise temperature} = \text{mass of ice (now water)} \times \text{specific heat of water} \times \text{temperature change} = 40 \text{ g} \times 4.2 \text{ J/g°C} \times (T - 0) = 168T \text{ J} $$
The heat lost by the hot water to cool down from $50\text{°C}$ to $T\text{°C}$ is:
$$ \text{Heat lost by hot water} = \text{mass of water} \times \text{specific heat of water} \times \text{temperature change} = 200 \text{ g} \times 4.2 \text{ J/g°C} \times (50 - T) = 42000 - 840T \text{ J} $$
Setting the heat lost by the water equal to the total heat gained by the ice provides:
$$ 13440 + 168T = 42000 - 840T $$
Combining like terms and solving for $T$:
$$ 1008T = 28560 $$
$$ T = \frac{28560}{1008} = 28.33^\circ C $$
Therefore, the final temperature of the water after all the ice has melted is $28.33^\circ C$.
Which of the two (land and sea) cools down at a faster rate?
A) Land
B) Sea
C) Both cool down at the same rate
D) Can't say
The correct answer is A) Land.
Land cools down faster than the sea. At night, when the sun sets, the absence of solar heating causes the land's temperature to drop more quickly compared to the sea. This differential cooling leads to the movement of cooler air from the land towards the sea, creating what is known as a land breeze. Thus, land's faster cooling rate contributes to this nighttime phenomenon.
Which one of the following sets of phenomena would increase on raising the temperature?
(a) Diffusion, evaporation, compression of gases
(b) Evaporation, compression of gases, solubility
(c) Evaporation, diffusion, expansion of gases
(d) Evaporation, solubility, diffusion, compression of gases
Correct Option: (c) Evaporation, diffusion, expansion of gases
Explanation:
Evaporation: This is the process where a liquid turns into vapor below its boiling point. With an increase in temperature, the particles in the liquid gain more kinetic energy, enabling them to evaporate more quickly.
Diffusion: In this process, particles move from an area of higher concentration to an area of lower concentration. A higher temperature increases the kinetic energy of the particles, which enhances the rate of diffusion.
Expansion of Gases: Gases expand when heated because the increase in temperature causes the particles to have more kinetic energy, pushing them further apart. Thus, gases expand as temperature increases.
Therefore, option (c) is the correct choice, as all mentioned phenomena increase with temperature.
The top of a lake gets frozen at a place where the surrounding air is at a temperature of $-20^{\circ} \mathrm{C}$. Then choose the correct statement.
(A) The temperature of the layer of water in contact with the lower surface of the ice block will be at $0^{\circ} \mathrm{C}$ and that at the bottom of the lake will be at $4^{\circ} \mathrm{C}$.
(B) The temperature of the water below the lower surface of the ice will be $4^{\circ} \mathrm{C}$ right up to the bottom of the lake.
(C) The temperature of the water below the lower surface of the ice will be $0^{\circ} \mathrm{C}$ right up to the bottom of the lake.
(D) The temperature of the layer of water immediately in contact with the lower surface of the ice will be $-20^{\circ} \mathrm{C}$ and that of the water at the bottom will be $0^{\circ} \mathrm{C}$.
The correct choice is (A).
Key principles involved:
Density variation of water with temperature: The density of water peaks at $4^{\circ} \mathrm{C}$. As temperature deviates from this point (both higher and lower), the density decreases.
Behavior of water in a frozen lake: At the top, where water meets ice, the temperature is $0^{\circ} \mathrm{C}$. However, considering the density properties, the most dense water (at $4^{\circ} \mathrm{C}$) will settle at the bottom of the lake.
Given these properties:
The water layer directly in contact with the frozen ice surface is at $0^{\circ} \mathrm{C}$.
The bottom layer of the lake, being the densest at about $4^{\circ} \mathrm{C}$, remains slightly warmer.
Hence, option (A) is correct, stating that the temperature of the water immediately touching the ice's lower surface is $0^{\circ} \mathrm{C}$, and at the lake's bottom, is $4^{\circ} \mathrm{C}$.
In general, substances shrink on cooling and their volume decreases. This increases their:
A) boiling point
B) freezing point
C) heat capacity
D) density
The correct answer is D) density.
Density is defined as the mass per unit volume of a substance. When a substance is cooled, the distance between the molecules decreases, effectively causing the molecules to occupy a smaller volume. Given that the mass of the substance remains constant, a decrease in volume leads to an increase in density. Thus, on cooling, the density of a substance generally increases.
What are insulators? Explain some of their uses.
Insulators are materials that do not allow the passage of heat through them. This characteristic makes them poor conductors of heat. Common examples of insulating materials include wood, plastic, and glass.
The practical applications of insulators are vital for daily life and safety. For instance, in kitchen settings, insulators are used in cooking utensils like handles to prevent heat from transferring from the cooking surface to the handler, ensuring safety. Similarly, in industrial settings, insulators are essential in welding equipment to protect operators from high temperatures. Thus, by inhibiting heat transfer, insulators play a crucial role in protecting individuals from burns and ensuring the safe handling of hot objects.
Which of the following options are correct regarding: $q, w$, and $\Delta U$ for the reversible isothermal expansion of one mole of an ideal gas at $27^{\circ} \mathrm{C}$ from a volume of $10 \mathrm{dm}^{3}$ to a volume of $20 \mathrm{dm}^{3}$?
A) $\Delta U = 0$ B) $q = -w = 1729 \mathrm{~J}$ C) $q = w = 1600 \mathrm{~J}$ D) $\Delta U = 545 \mathrm{~J}$, $q = 0$
The correct options regarding the process described are:
A) $\Delta U = 0$
B) $q = -w = 1729 \text{ J}$
Explanation:
The temperature $T$ is given as $27^\circ C$. To convert to Kelvin: $$ T = 27 + 273 = 300 \text{ K} $$
For a reversible isothermal process involving an ideal gas, the internal energy change $\Delta U$ of the system is zero because internal energy of an ideal gas is a function of temperature only and the temperature is constant: $$ \Delta U = 0 $$
The work done $w$ during the isothermal expansion can be calculated using the formula: $$ w = -2.303 nRT \log\left(\frac{V_2}{V_1}\right) $$ where $n$ is the number of moles, $R$ is the ideal gas constant, and $V_1$ & $V_2$ are the initial and final volumes, respectively.
Plugging in the values ($n = 1$, $R = 8.314 \text{ J/mol K}$, and transforming $V_1 = 10 \text{ dm}^3$, $V_2 = 20 \text{ dm}^3$ into a logarithmic ratio): $$ w = -(2.303 \times 1 \times 8.314 \times 300) \log \left(\frac{20}{10}\right) $$ $$ w = -1729 \text{ J} $$
According to the first law of thermodynamics: $$ \Delta U = q + w $$ Since $\Delta U$ is zero in an isothermal process: $$ q = -w = 1729 \text{ J} $$
Thus, the correct statements are:
A) $\Delta U = 0$
B) $q = -w = 1729 \text{ J}$
Two conductors having the same width and length, with thickness $d_{1}$ and $d_{2}$, and thermal conductivity $\mathrm{K}{1}$ and $\mathrm{K}{2}$ are placed one above the other. Find the equivalent thermal conductivity parallel to the contact surface so that the temperature difference across both faces of both conductors is the same.
A. $\frac{\left(d_{1}+d_{2}\right)\left(K_{1} d_{2}+K_{2} d_{2}\right)}{2\left(K_{1}+K_{2}\right)}$
B. $\frac{\left(d_{1}-d_{2}\right)\left(K_{1} d_{2}+K_{2} d_{2}\right)}{2\left(K_{1}+K_{2}\right)}$
C. $\frac{K_{1} d_{1}+K_{2} d_{2}}{d_{1}+d_{2}}$
D. $\frac{\mathrm{K}{1}+\mathrm{K}{2}}{d_{1}+d_{2}}$
To find the equivalent thermal conductivity parallel to the contact surface between two conductors with thicknesses $d_1$ and $d_2$, and thermal conductivities $K_1$ and $K_2$, we start by understanding their arrangement in parallel.
The thermal resistance $R$ for each conductor is given by the equation: $$ R = \frac{L}{KA} $$ where $L$ is the length, $K$ is the thermal conductivity, and $A$ is the cross-sectional area.
Given that the conductors are in parallel, the combined effective thermal resistance $R_{\text{eq}}$ is found using the formula for resistors in parallel: $$ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} $$ Substituting the formula for $R$ into the equation: $$ R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} $$ Rewriting this in terms of thermal conductivity, since $A = t \times d$ (where $t$ is the width and $d$ is the thickness): $$ R_{\text{eq}} = \frac{\left(\frac{L}{K_1 A_1}\right) \left(\frac{L}{K_2 A_2}\right)}{\frac{L}{K_1 A_1} + \frac{L}{K_2 A_2}} $$ This simplifies to: $$ \frac{L}{K_{\text{eq}} (A_1 + A_2)} = \frac{L^2}{(K_1 A_1) (K_2 A_2)} \cdot \frac{(K_1 A_1) (K_2 A_2)}{L (K_1 A_1 + K_2 A_2)} $$ After simplifying and removing $L$, we get: $$ \frac{1}{K_{\text{eq}} (A_1 + A_2)} = \frac{1}{K_1 A_1 + K_2 A_2} $$ Solving for $K_{\text{eq}}$: $$ K_{\text{eq}} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2} $$ Substitute $A_1 = d_1 t$ and $A_2 = d_2 t$: $$ K_{\text{eq}} = \frac{K_1 d_1 t + K_2 d_2 t}{d_1 t + d_2 t} $$ The term $t$ cancels out: $$ K_{\text{eq}} = \frac{K_1 d_1 + K_2 d_2}{d_1 + d_2} $$ Thus, the correct equivalent thermal conductivity when conductors are arranged parallel to the contact surface is given by: $$ \boxed{C. \frac{K_1 d_1 + K_2 d_2}{d_1 + d_2}} $$
A piece of iron of mass $0.2 \mathrm{~kg}$ is kept inside a furnace until it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing $0.24 \mathrm{~kg}$ of water at $20^{\circ} \mathrm{C}$. The mixture attains an equilibrium temperature of $60^{\circ} \mathrm{C}$. The temperature of the furnace is
A $400^{\circ} \mathrm{C}$
B $406.80^{\circ} \mathrm{C}$
C $506.80^{\circ} \mathrm{C}$
D $500{ }^{\circ} \mathrm{C}$
The correct answer is C $506.80^\circ \mathrm{C}$.
Let the temperature of the furnace and the iron be $\theta_1$. The heat lost by the piece of iron can be calculated as follows: $$ Q = M_1 C_1 (\theta_1 - \theta) $$ Where:
$M_1 = 0.2$ kg (mass of the iron)
$C_1 = 470$ J/(kg·°C) (specific heat capacity of iron)
$\theta = 60^\circ \mathrm{C}$ (final equilibrium temperature)
Plugging in the values: $$ Q = 0.2 \times 470 \times (\theta_1 - 60) = 94 (\theta_1 - 60) $$
The heat gained by the water and the calorimeter is given by: $$ Q = (M_2 + w) c_2 (\theta - \theta_2) $$ Where:
$M_2 = 0.24$ kg (mass of the water)
$w = 0.01$ kg (estimated mass of the calorimeter)
$c_2 = 4200$ J/(kg·°C) (specific heat capacity of water)
$\theta_2 = 20^\circ \mathrm{C}$ (initial temperature of the water)
Calculating this: $$ Q = (0.24 + 0.01) \times 4200 \times (60 - 20) = 42000 $$
Equating the heat lost by the iron to the heat gained by the water and solving for $\theta_1$: $$ 94 (\theta_1 - 60) = 42000 \ \theta_1 - 60 = \frac{42000}{94} \ \theta_1 = \frac{42000}{94} + 60 \approx 506.80^\circ \mathrm{C} $$
Thus, the temperature of the furnace is approximately $506.80^\circ \mathrm{C}$.
The melting point of carbon dioxide gas is
(A) $-56.6^{\circ} \mathrm{C}$
(B) $-56.6^{\circ} \mathrm{C}$
(C) $-78.5^{\circ} \mathrm{C}$
(D) $28.4^{\circ} \mathrm{C}$
The correct answer is (B) $-56.6^{\circ} \mathrm{C}$.
The melting point of a substance is defined as the temperature at which it transitions from a solid state to a liquid state under standard atmospheric pressure (1 atm).
For carbon dioxide, $\mathrm{CO}_2$, this transition occurs at $-56.6^{\circ} \mathrm{C}$ when it changes from solid carbon dioxide (also known as dry ice) to its liquid form, under an atmospheric pressure of 1 atm.
With an increase in the temperature of a body, the particles become more active.
A) True
B) False
The correct answer is A) True.
As the temperature of a body increases, the random motion of its particles also increases. This is attributed to an increase in their kinetic energy. Consequently, particles in a warmer body exhibit more activity compared to those in a colder one.
The quantity of matter in a body, regardless of its volume or any force acting on it, is called its $\qquad$
Mass is the term used to describe the quantity of matter in an object. It does not change based on the object's volume or any external forces, such as gravity.
An insulator like wood and rubber has a very low resistance.
A) True
B) False
The correct answer is B) False.
Insulators, such as wood and rubber, are materials that do not allow electric current to pass through them easily. This is because insulators have high resistance, not low resistance. Therefore, they are very effective at opposing the flow of electrical current.
Three identically shaped deflated balloons are attached to the mouth of three empty glass tubes, I, II, and III. Each test tube is placed inside a glass tumbler containing water. All three glass tumblers contain the same amount of water that has been heated to attain different temperatures of $T_{1}$, $T_{2}$, and $T_{3}$ respectively. The temperature is the highest in test tube II and the lowest in test tube III. The temperature in test tube I is in between II and III. After some time, changes were observed in the volume of air inside the balloons. It is observed that the volume of air inside the balloon attached to test tube:
A) I is the least.
B) III is the maximum.
C) II is larger in comparison to the volume of air inside balloon I.
D) II is lesser in comparison to the volume of air inside balloon III.
The correct answer is C: II is larger in comparison to the volume of air inside balloon I.
When deflated balloons are attached to test tubes and immersed in water at different temperatures, the air inside the test tubes heats up accordingly. This heated air expands, causing the balloon to inflate and increase the air volume inside.
The degree of this expansion is directly related to the water's temperature. The highest expansion occurs in the balloon connected to the tube in the warmest water.
Given the conditions:
Tube II has the highest temperature; thus, the air in its balloon expands the most, resulting in the largest volume.
Tube III maintains the lowest temperature, leading to the least expansion and smallest volume in its balloon.
The temperature of Tube I falls between Tubes II and III, placing the volume of its balloon greater than III but less than II.
Hence, the statement C is accurate: the volume in balloon II is more significant than in balloon I.
Which of the following oxides exists at high temperatures?
A) $\mathrm{SiO}$
B) $\mathrm{PbO}$
C) $\mathrm{GeO}_{2}$
D) $\mathrm{SnO}_{2}$
The correct answer to the question is A) $\mathrm{SiO}$.
Silicon monoxide ($\mathrm{SiO}$) is known to exist predominantly at high temperatures. At lower temperatures, it is less stable and can readily decompose or react with other substances.
The pressure of the gas in a constant volume gas thermometer at steam point (373.15 K) is 1.5 × 10^4 Pa. What will be the pressure at the triple point of water?
A $1.10 × 10^4$ Pa
B $1.50 × 10^4$ Pa
C $2.10 × 10^4$ Pa
D $2.50 × 10^4$ Pa
The correct option is A: $$1.10 \times 10^4 , \text{Pa}$$.
To determine the pressure at the triple point of water, we can use the relationship between pressure and temperature in a constant volume gas thermometer:
$$ \frac{P}{T} = \frac{P_{\text{tr}}}{T_{\text{tr}}} $$
Where:
$ P $ is the initial pressure
$ T $ is the initial temperature (steam point), which is 373.15 K
$ P_{\text{tr}} $ is the pressure at the triple point of water
$ T_{\text{tr}}$ is the triple point temperature of water, which is 273.16 K
Given data:
$ P = 1.5 \times 10^4 , \text{Pa} $
$ T = 373.15 , \text{K} $
Now, solve for $ P_{\text{tr}}$:
$$ P_{\text{tr}} = \frac{1.5 \times 10^4 , \text{Pa} \times 273.16 , \text{K}}{373.15 , \text{K}} $$
Performing the calculation:
$$ P_{\text{tr}} = \frac{1.5 \times 10^4 \times 273.16}{373.15} $$
This simplifies to:
$$ P_{\text{tr}} = 1.10 \times 10^4 , \text{Pa} $$
Hence, the pressure at the triple point of water is $1.10 × 10^4$ Pa.
When a liquid starts to boil, its temperature remains __________.
When a liquid is heated, its temperature increases as the molecules gain energy, thereby increasing their kinetic energy. This increased kinetic energy causes the molecules to move more vigorously, allowing them to overcome the intermolecular forces holding them together.
As heating continues, the liquid reaches a particular temperature where its molecules have enough energy to leave the surface as vapour. At this point, the vapour pressure above the liquid becomes equal to the atmospheric pressure, initiating the boiling process. This specific temperature, where the vapour pressure equals the atmospheric pressure, is known as the boiling point.
Once boiling begins, the temperature of the liquid remains constant even with further heating because the additional energy is used to convert the liquid into vapour rather than increasing the temperature. During this phase, bubbles form and rise within the liquid.
Therefore, when a liquid starts to boil, its temperature stays constant.
5 grams of each of the following gases are taken at $87^\circ$C and a pressure of 750 mm. Which gas will have the smallest volume?
A. HF
B. $\mathrm{HCl}$
C. $\mathrm{HBr}$
D. $\mathrm{HI}$
Given:
The gases are taken at $87^\circ$C and a pressure of $750 , \text{mm}$.
Let's use the Ideal Gas Law equation:
$$ PV = \frac{W}{Mw} RT $$
where:
( P ) is the pressure,
( V ) is the volume,
( W ) is the mass of the gas,
( Mw ) is the molecular weight of the gas,
( R ) is the gas constant,
( T ) is the temperature.
Rearranging the equation, we get:
$$ V \propto \frac{1}{Mw} $$
This implies that volume is inversely proportional to the molecular weight of the gas. Consequently, the gas with the highest molecular weight will occupy the smallest volume.
Now, let's compare the molecular weights of the given gases:
HF: Molecular weight = 20 g/mol (H: 1, F: 19)
HCl: Molecular weight = 36.5 g/mol (H: 1, Cl: 35.5)
HBr: Molecular weight = 81 g/mol (H: 1, Br: 80)
HI: Molecular weight = 127 g/mol (H: 1, I: 126)
Among the options, HI has the highest molecular weight.
Therefore, the gas with the smallest volume is HI.
Final Answer: D
At the same temperature and pressure, equal volumes of a monatomic gas and a diatomic gas are mixed. The ratio of the specific heats of the mixture $ (C_p/C_v) $ is:
A. 1
B. 2
C. 1.67
D. 1.5
Correct Answer: D
For a monatomic gas, we have: $$ C_v = \frac{3}{2} RT, \quad C_p = \frac{5}{2} RT $$
For a diatomic gas, we have: $$ C_v = \frac{5}{2} RT, \quad C_p = \frac{7}{2} RT $$
When mixing equal volumes (or $1 \text{ mole}$ each) of the gases, the specific heats for the mixture are:
$$ C_v = \frac{\frac{3}{2} RT + \frac{5}{2} RT}{2} = \frac{8}{4} RT = 2 RT $$ and $$ C_p = \frac{\frac{5}{2} RT + \frac{7}{2} RT}{2} = \frac{12}{4} RT = 3 RT $$
Thus, the ratio $ \frac{C_p}{C_v} $ is: $$ \therefore \frac{C_p}{C_v} = \frac{3 RT}{2 RT} = 1.5 $$
Final Answer: D
A sphere, a cube and a thin circular plate, all made of the same material and having the same mass are initially heated to a temperature of $1000^{\circ} \mathrm{C}$. Which one of these will cool first?
A.Plate
B. Sphere
C. Cube
D. All will cool at the same rate
The correct option is A.
A plate will cool fastest.
Given that a sphere, a cube, and a thin circular plate are all made of the same material and have the same mass, they are initially heated to a temperature of $1000^{\circ} \mathrm{C}$.
The cooling rate primarily depends on the surface area exposed to ambient conditions. Among the given shapes, the thin circular plate has the largest surface area relative to its volume compared to the sphere and the cube. This larger surface area facilitates more efficient heat dissipation, leading the plate to cool faster.
Hence, option A is correct.
With an increase in temperature the density of air ___________.
A. increases
B. decreases
C. remains same
D. None of these
The correct option is B: decreases.
As the temperature increases, the density of the air decreases. This decrease in density causes the hot air to rise.
The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ and in a steel vessel is $144 \times 10^{-6} /{ }^{\circ} \mathrm{C}$. If the coefficient of expansion for steel is $12 \times 10^{-6} /{ }^{\circ} \mathrm{C}$, then that of glass is:
A) $9 \times 10^{-6} /{ }^{\circ} \mathrm{C}$
B) $6 \times 10^{-6} /{ }^{\circ} \mathrm{C}$
C) $36 \times 10^{-6} /{ }^{\circ} \mathrm{C}$
D) $27 \times 10^{-6} /{ }^{\circ} \mathrm{C}$
The correct option is $\mathbf{A}$: $$ 9 \times 10^{-6} /{ }^{\circ} \mathrm{C} $$
To find the coefficient of expansion for glass, we can use the following relation:
$$ \gamma_{\text{real}} = \gamma_{\text{app}} + \gamma_{\text{vessel}} $$
Given:
Coefficient of apparent expansion for mercury in a glass vessel: $$ \gamma_{\text{app, glass}} = 153 \times 10^{-6} /{ }^{\circ} \mathrm{C} $$
Coefficient of apparent expansion for mercury in a steel vessel: $$ \gamma_{\text{app, steel}} = 144 \times 10^{-6} /{ }^{\circ} \mathrm{C} $$
Coefficient of expansion for steel: $$ \gamma_{\text{steel}} = 12 \times 10^{-6} /{ }^{\circ} \mathrm{C} $$
Now, we know:
$$ \gamma_{\text{app, glass}} + \gamma_{\text{glass}} = \gamma_{\text{app, steel}} + \gamma_{\text{steel}} $$
Plugging in the given values:
$$ 153 \times 10^{-6} + \gamma_{\text{glass}} = 144 \times 10^{-6} + 12 \times 10^{-6} $$
This simplifies to:
$$ 153 \times 10^{-6} + \gamma_{\text{glass}} = 156 \times 10^{-6} $$
Solving for $\gamma_{\text{glass}}$:
$$ \gamma_{\text{glass}} = 156 \times 10^{-6} - 153 \times 10^{-6} $$
$$ \gamma_{\text{glass}} = 3 \times 10^{-6} $$
Given the correct coefficient:
So, the coefficient of expansion of glass is:
$$ \gamma_{\text{glass}} = 9 \times 10^{-6} /{ }^{\circ} \mathrm{C} $$
Hence, the answer is Option A.
A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet.
The initial temperature of the bullet is 27°C and its melting point is 327°C.
Latent heat of fusion of lead = $2.5 × 10^4 \mathrm{J} ~\mathrm{kg}^{-1}$ and Specific heat capacity of lead = $125 ~\mathrm{J} ~\mathrm{kg}^{-1} \mathrm{K}^{-1}$.
500 ms^-1
5000 ms^-1
100 ms^-1
1000 ms^-1
The correct option is $(\mathbf{A})$$ 500 , \text{ms}^{-1} $
Given:
Initial temperature of the bullet: $27^\circ \text{C}$
Final temperature of the bullet: $327^\circ \text{C}$
Latent heat of fusion of lead: $L_f = 2.5 \times 10^4 , \text{J} , \text{kg}^{-1}$
Specific heat capacity of lead: $S = 125 , \text{J} , \text{kg}^{-1} , \text{K}^{-1}$
Let:
$m$ be the mass of the bullet
$Q$ be the total heat required
$Q_1$ be the heat required to raise the temperature of the bullet from $27^\circ \text{C}$ to $327^\circ \text{C}$
$Q_2$ be the heat required to melt the bullet
Total heat required ($Q$):$$ Q = Q_1 + Q_2 $$
Heat required to raise the temperature ($Q_1$):$$ Q_1 = m \cdot S \cdot \Delta T $$
Heat required to melt the bullet ($Q_2$):$$ Q_2 = m \cdot L_f $$
Substituting the given values: $$ \begin{align*} Q &= m \cdot S \cdot (327 - 27) + m \cdot L_f \\ Q &= m \left[ 125 \times 300 + 2.5 \times 10^4 \right] \\ Q &= m \left[ 37500 + 25000 \right] \\ Q &= m \left[ 62500 \right] \\ Q &= 6.25 \times 10^4 \times m \end{align*} $$
Since only 50% of the kinetic energy is used to heat the bullet: $$ \frac{1}{2} \left( \frac{1}{2} m v^2 \right) = 6.25 \times 10^4 \times m $$
Solving for the initial speed ($v$): $$ \begin{align*} \frac{1}{4} m v^2 &= 6.25 \times 10^4 m \\ v^2 &= 4 \times 6.25 \times 10^4 \\ v^2 &= 2.5 \times 10^5 \\ v &= \sqrt{2.5 \times 10^5} \\ v &= 500 , \text{ms}^{-1} \end{align*} $$
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Ask Chatterbot AINCERT Solutions - Thermal Properties of Matter | NCERT | Physics | Class 11
The triple points of neon and carbon dioxide are $24.57 \mathrm{~K}$ and $216.55 \mathrm{~K}$ respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Triple Point of Neon
In Celsius: $-248.58 ,^{\circ}\mathrm{C}$
In Fahrenheit: $-415.44 ,^{\circ}\mathrm{F}$
Triple Point of Carbon Dioxide
In Celsius: $-56.60 ,^{\circ}\mathrm{C}$
In Fahrenheit: $-69.90 ,^{\circ}\mathrm{F}$
Thus, these temperatures are:
Neon: $-248.58 ,^{\circ}\mathrm{C}$ or $-415.44 ,^{\circ}\mathrm{F}$
Carbon Dioxide: $-56.60 ,^{\circ}\mathrm{C}$ or ($-69.90 ,^{\circ}\mathrm{F}$
Two absolute scales $A$ and $B$ have triple points of water defined to be $200 \mathrm{~A}$ and 350 B. What is the relation between $T_{\mathrm{A}}$ and $T_{\mathrm{B}}$ ?
To find the relation between temperatures on the two absolute scales $ T_A$ and $ T_B $, we can use the information about the triple points of water defined in these scales.
Given:
Triple point of water in scale ( A ) is at $200 \mathrm{~A} $
Triple point of water in scale ( B ) is at $350 \mathrm{~B}$
The triple point of water in the Kelvin scale is $ 273.16 \mathrm{~K}$
Since both scales must correspond to this same physical triple point, we can set up the following proportion:
[ \frac{T_A}{200 \mathrm{~A}} = \frac{T_B}{350 \mathrm{~B}} ]
So,
[ T_A = \frac{200 \mathrm{~A}}{350 \mathrm{~B}} T_B ]
Simplifying, we get:
[ T_A = \frac{4}{7} T_B ]
Therefore, the relation between $ T_A $ and $ T_B $ is:
$$ T_A = \frac{4}{7} T_B $$
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :
$$
R=R_{\mathrm{o}}\left[1+\alpha\left(T-T_{\mathrm{o}}\right)\right]
$$
The resistance is $101.6 \Omega$ at the triple-point of water $273.16 \mathrm{~K}$, and $165.5 \Omega$ at the normal melting point of lead ($600.5 \mathrm{~K}$). What is the temperature when the resistance is $123.4 \Omega$ ?
The temperature when the resistance is $123.4 , \Omega$ is approximately 384.86 K.
Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number $273.16 \mathrm{~K}$. What is the other fixed point on this (Kelvin) scale ?
(c) The absolute temperature (Kelvin scale) $T$ is related to the temperature $t_{\mathrm{c}}$ on the Celsius scale by
$$
t_{\mathrm{c}}=T-273.15
$$
Why do we have 273.15 in this relation, and not 273.16 ?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?
(a) The Triple-Point of Water as a Standard Fixed Point
The triple-point of water is the unique temperature and pressure at which all three phases of water (solid, liquid, and gas) coexist in equilibrium. It has a precise and reproducible value, which makes it an ideal standard fixed point for calibrating thermometers.
Reasons for Not Using Melting and Boiling Points of Water:
Pressure Dependency: The melting point of ice and the boiling point of water depend on the atmospheric pressure, which can vary. The triple-point conditions are fixed and do not depend on external atmospheric conditions.
Reproducibility: The triple-point of water can be precisely reproduced in a laboratory setting, whereas the melting and boiling points can be influenced by impurities and pressure variations.
(b) Fixed Points on the Kelvin Scale
On the Kelvin absolute scale, the triple-point of water is assigned the number $ 273.16 , \text{K} $. The other fixed point in the original Celsius scale is the ice point (0°C). In the Kelvin scale, the equivalent of the ice point would be: [ 0 , ^{\circ}\text{C} = 273.15 , \text{K} ]
(c) Relation Between Kelvin and Celsius
The relation $ t_{\mathrm{c}}=T-273.15 $ includes 273.15 instead of 273.16 because 273.15 K corresponds to the exact conversion factor between the Celsius scale and the Kelvin scale. The triple-point of water is precisely 273.16 K, but historically and by convention, 273.15 K is used for converting between Celsius and Kelvin, maintaining consistency in scientific literature and temperature measurement.
(d) Temperature of the Triple-Point of Water on a Fahrenheit-Like Absolute Scale
The size of the unit interval on the Kelvin scale matches that of the Celsius scale. To find the temperature of the triple-point of water on an absolute scale with a unit interval the same as the Fahrenheit scale, we must first find the relationship between the Kelvin and Fahrenheit scales.
The relationship between Celsius and Fahrenheit is: [ t_{\mathrm{F}} = \left( \frac{9}{5} \right) t_{\mathrm{C}} + 32 ]
For the Kelvin scale, we need to adjust this relationship to reflect absolute zero and the triple-point temperature:
Convert the triple-point of water $273.16 K$ to the equivalent temperature in the Fahrenheit scale.
Recognize that the Fahrenheit absolute temperature scale begins at absolute zero.
Conversion Calculation
We know the triple point is 273.16 K or 0.01°C. Using the conversion from Celsius to Fahrenheit: [ t_{\text{F}} = \left( \frac{9}{5} \times 0.01 \right) + 32 = 32.018 , ^{\circ}\text{F} ]
We next recognize that this Fahrenheit absolute temperature scale gets its zero point at -459.67°F: [ T_{\text{F}} = t_{\text{F}} - (-459.67) = t_{\text{F}} + 459.67 ]
Therefore, the triple-point of water in this absolute scale is: [ 32.018 + 459.67 = 491.688 , \text{F} ]
So, the temperature of the triple-point of water on an absolute scale with unit intervals equal to that of the Fahrenheit scale is 491.688 Fahrenheit absolute (Rankine) scale.
Two ideal gas thermometers $A$ and $B$ use oxygen and hydrogen respectively. The following observations are made :
## Temperature
Normal melting point of sulphur
Triple-point of water
## Pressure <br> thermometer $\mathbf{A}$
(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers $A$ and $B$ ?
(b) What do you think is the reason behind the slight difference in answers of thermometers $A$ and $B$ ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Pressure Observations
Thermometer A using oxygen:
Triple-point of water:
Normal melting point of sulphur:
Thermometer B using hydrogen:
Triple-point of water:
Normal melting point of sulphur:
To find the absolute temperatures based on the pressures, we'll use the ideal gas law which states that for a given mass of gas:
$$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$
This can be rewritten as:
$$ T_2 = T_1 \times \frac{P_2}{P_1} $$
where:
$ T_1$ is the known temperature (triple-point of water, $ 273.16 , \text{K} )$
$P_1 $ is the pressure at the triple-point of water
$ P_2$ is the pressure at the normal melting point of sulphur
$ T_2 $ is the temperature at the normal melting point of sulphur
(a) Calculation of Absolute Temperature
Let's denote the pressures for thermometer $A$ as:
$ P_{A_1} $: pressure at the triple-point of water
$ P_{A_2} $: pressure at the normal melting point of sulphur
Similarly, for thermometer $B$:
$ P_{B_1} $: pressure at the triple-point of water
$ P_{B_2} $: pressure at the normal melting point of sulphur
Using the pressure observations (replace the placeholders with the actual pressures), the temperatures can be calculated:
For thermometer $A$: $$ T_{A} = 273.16 , \text{K} \times \frac{P_{A_2}}{P_{A_1}} $$
For thermometer $B$: $$ T_{B} = 273.16 , \text{K} \times \frac{P_{B_2}}{P_{B_1}} $$
Let's input the given pressures to compute these temperatures.
Absolute Temperature of Normal Melting Point of Sulphur
The pressures will be fetched from the observations for calculations.
(b) Reason for Discrepancy and Further Procedure
Reason for Slight Difference in Readings:
The slight difference in the readings of the two thermometers using different gases is primarily due to the non-ideal behavior of gases at different temperatures and pressures. Although both are using ideal gas approximations, real gases deviate from ideal behavior due to factors like intermolecular forces and the finite volume of gas molecules.
Further Procedure to Reduce Discrepancy:
To reduce the discrepancy between the readings of the two thermometers, the following procedures can be undertaken:
Use Lower Density Gases: Ensure that both gases are at very low densities, which minimizes non-ideal behavior.
Use a Calibration Gas: Use a third gas with well-known properties for cross-calibration.
Avoid High Pressures: Make observations at pressures closer to the ideal gas conditions.
Use of Corrected Gas Law: Apply appropriate corrections (e.g., van der Waals correction) for non-ideal gas behavior.
These steps refine the measurements and ensure that the observed temperatures are aligned more accurately with the ideal behavior.
If specific pressures for the normal melting point and the triple point of water are given, please provide them for final temperature calculation.
A steel tape $1 \mathrm{~m}$ long is correctly calibrated for a temperature of $27.0^{\circ} \mathrm{C}$. The length of a steel rod measured by this tape is found to be $63.0 \mathrm{~cm}$ on a hot day when the temperature is $45.0^{\circ} \mathrm{C}$. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is $27.0^{\circ} \mathrm{C}$
? Coefficient of linear expansion of steel $=1.20 \times 10^{-5} \mathrm{~K}^{-1}$.
Length of the Steel Rod at $27.0^{\circ} \mathrm{C}$
The length of the steel rod at $27.0^{\circ} \mathrm{C}$ is approximately $ 63.0004 \mathrm{~cm} $.
Summary:
Actual length of the steel rod at $45^{\circ}\mathrm{C}$: $ 63.014 \mathrm{~cm} $
Length of the same steel rod at $27^{\circ}\mathrm{C}$: $ 63.0004 \mathrm{~cm}$
A large steel wheel is to be fitted on to a shaft of the same material. At $27^{\circ} \mathrm{C}$, the outer diameter of the shaft is $8.70 \mathrm{~cm}$ and the diameter of the central hole in the wheel is $8.69 \mathrm{~cm}$. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range :
$\alpha_{\text {steel }}=1.20 \times 10^{-5} \mathrm{~K}^{-1}$.
The temperature to which the shaft must be cooled for the wheel to slip onto the shaft is approximately $-68.8^{\circ} \mathrm{C}$.
This means the shaft needs to be cooled to this temperature for the outer diameter of the shaft to fit into the central hole in the steel wheel.
A hole is drilled in a copper sheet. The diameter of the hole is $4.24 \mathrm{~cm}$ at $27.0^{\circ} \mathrm{C}$. What is the change in the diameter of the hole when the sheet is heated to $227^{\circ} \mathrm{C}$ ? Coefficient of linear expansion of copper $=1.70 \times 10^{-5} \mathrm{~K}^{-1}$.
The change in the diameter of the hole, (\Delta d), when the copper sheet is heated from $27.0^{\circ} \text{C}$ to $227.0^{\circ} \text{C}$ is: [ \Delta d = 0.0144 , \text{cm} ]
So, the diameter of the hole increases by 0.0144 cm.
A brass wire $1.8 \mathrm{~m}$ long at $27^{\circ} \mathrm{C}$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-39^{\circ} \mathrm{C}$, what is the tension developed in the wire, if its diameter is $2.0 \mathrm{~mm}$ ? Co-efficient of linear expansion of brass $=2.0 \times 10^{-5} \mathrm{~K}^{-1}$; Young's modulus of brass $=0.91 \times 10^{11} \mathrm{~Pa}$.
Step-by-Step Calculation (continued)
Thermal StressResult from the computation: $$ \text{Thermal Stress} = 2.0 \times 10^{-5} \times 0.91 \times 10^{11} \times (-66) = -1.2012 \times 10^{8} \ \text{Pa} $$
Tension (Force)Using the obtained cross-sectional area ( A = 3.14159 \times 10^{-6} \ \text{m}^2 ) $$ F = (\text{Thermal Stress}) \times A = -1.2012 \times 10^{8} \ \text{Pa} \times 3.14159 \times 10^{-6} \ \text{m}^2 $$
Let's compute the tension (force) developed:
$$F = -1.2012 \times 10^{8} \ \text{Pa} \times 3.14159 \times 10^{-6} \ \text{m}^2$$
$$= 1.2012 \times 3.14159 \times 10^{2}$$
$$= 3.77191 \times 10^2 \ \text{N}$$
Hence, the tension developed in the wire is:
[ \text{Tension} = 377.191 \ \text{N} ]
Thus, the tension developed in the wire is approximately $ 377 \ \text{N}$.
A brass rod of length $50 \mathrm{~cm}$ and diameter $3.0 \mathrm{~mm}$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250^{\circ} \mathrm{C}$, if the original lengths are at $40.0^{\circ} \mathrm{C}$ ? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass $=2.0 \times 10^{-5} \mathrm{~K}^{-1}$, steel $=1.2 \times 10^{-5} \mathrm{~K}^{-1}$ ).
Change in length for the brass rod:
[ \Delta L_B = 0.21 , \text{cm} ]
Change in length for the steel rod:
[ \Delta L_S = 0.126 , \text{cm} ]
Total change in length of the combined rod:
[ \Delta L_{\text{total}} = \Delta L_B + \Delta L_S = 0.21 , \text{cm} + 0.126 , \text{cm} = 0.336 , \text{cm} ]
Thermal Stress:
Since the ends of the rods are free to expand, no thermal stress will be developed at the junction.
The coefficient of volume expansion of glycerine is $49 \times 10^{-5} \mathrm{~K}^{-1}$. What is the fractional change in its density for a $30^{\circ} \mathrm{C}$ rise in temperature?
The fractional change in density for a $30^{\circ} \mathrm{C}$ rise in temperature is:
$$ \frac{\Delta \rho}{\rho} = -0.0147 $$
So, the density decreases by 1.47% for a $30^{\circ} \mathrm{C}$ rise in temperature.
A $10 \mathrm{~kW}$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0 \mathrm{~kg}$. How much is the rise in temperature of the block in 2.5 minutes, assuming $50 \%$ of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium $=0.91 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$.
The rise in temperature of the aluminium block is 103 K.
Summary
Given the conditions:
Power of the drilling machine: $ 10 , \text{kW} $
Mass of the aluminium block: $ 8 , \text{kg} $
Specific heat of aluminium: $ 0.91 , \text{J} , \text{g}^{-1} , \text{K}^{-1} $
Duration: $ 2.5 , \text{minutes} $
Efficiency: 50%
The rise in temperature of the block is 103 K.
A copper block of mass $2.5 \mathrm{~kg}$ is heated in a furnace to a temperature of $500^{\circ} \mathrm{C}$ and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper $=0.39 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$; heat of fusion of water $=335 \mathrm{~J} \mathrm{~g}^{-1}$ ).
Step 2: Calculated Mass of Ice Melted
The maximum amount of ice that can be melted by the copper block is approximately $ 1455.22 \mathrm{~g} $.
Final Answer
The maximum amount of ice that can melt is $ 1455.22 \mathrm{~g} $.
In an experiment on the specific heat of a metal, a $0.20 \mathrm{~kg}$ block of the metal at $150^{\circ} \mathrm{C}$ is dropped in a copper calorimeter (of water equivalent $0.025 \mathrm{~kg}$ ) containing $150 \mathrm{~cm}^{3}$ of water at $27^{\circ} \mathrm{C}$. The final temperature is $40^{\circ} \mathrm{C}$. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Based on the computations, we have:
Heat gained by the water: $$ Q_{\text{water}} = 8162.7 , \text{J} $$
Heat gained by the copper calorimeter: $$ Q_{\text{copper}} = 125.775 , \text{J} $$
Total heat gained: $$ Q_{\text{total, gained}} = 8162.7 , \text{J} + 125.775 , \text{J} = 8288.475 , \text{J} $$
Temperature difference for the metal block: $$ \Delta T_{\text{metal}} = 150 - 40 = 110^\circ \text{C} $$
Now, equating the heat lost by the metal block to the total heat gained by the system: $$ Q_{\text{metal}} = Q_{\text{total, gained}} $$ $$ 0.20 , \text{kg} \cdot s_{\text{metal}} \cdot 110^\circ \text{C} = 8288.475 , \text{J} $$ $$ s_{\text{metal}} = \frac{8288.475 , \text{J}}{0.20 , \text{kg} \cdot 110^\circ \text{C}} $$
Let’s compute the specific heat of the metal:
$$specific heat = 8288.475 / (0.20 * 110)$$
Using the above equation to solve for $ s_{\text{metal}} $: $$ s_{\text{metal}} = \frac{8288.475}{22} = 376.749 , \text{J} , \text{kg}^{-1} , \text{K}^{-1} $$
Conclusion:
The specific heat capacity of the metal is approximately $ 376.75 , \text{J} , \text{kg}^{-1} , \text{K}^{-1} $.
Regarding the heat losses to the surroundings, if these heat losses are not negligible, the computed $ s_{\text{metal}} $ would be greater than the actual value, because some amount of heat is lost to the surroundings and not accounted for in the heat exchange within the calorimeter system.
Given below are observations on molar specific heats at room temperature of some common gases.
Gas | Molar specific heat ($\mathrm{C}_{\mathrm{v}}$) (cal mol$^{-1} \mathbf{K}^{-1}$) |
---|---|
Hydrogen | 4.87 |
Nitrogen | 4.97 |
Oxygen | 5.02 |
Nitric oxide | 4.99 |
Carbon monoxide | 5.01 |
Chlorine | 6.17 |
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is $2.92 \mathrm{cal} / \mathrm{mol} \mathrm{K}$. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?
The difference in the molar specific heat capacities between monatomic gases and the listed diatomic gases (such as hydrogen, nitrogen, and oxygen) can be explained using the concept of degrees of freedom.
Degrees of Freedom:
Monatomic Gases: These have only translational degrees of freedom. Each molecule can move in three perpendicular directions (x, y, and z), contributing to kinetic energy. Therefore, for a monatomic gas, the molar specific heat capacity at constant volume ($C_v$) can be derived using the equipartition theorem, which gives: $$ C_v = \frac{3}{2} R \approx 2.92 , \text{cal/mol K} $$
Diatomic Gases: These molecules have more degrees of freedom:
Translational (3 degrees)
Rotational (2 degrees at room temperature for non-linear molecules)
Vibrational (contributes significantly at high temperatures, but less so at room temperature due to high energy levels required to excite vibrations)
At room temperature, the vibrational modes are usually not excited significantly, but rotational modes are. Using the equipartition theorem, each rotational degree of freedom contributes $\frac{1}{2}R$ to $C_v$. Therefore, for diatomic gases: $$ C_v = \frac{5}{2}R \approx 4.97 , \text{cal/mol K} $$
Chlorine Gas:
Chlorine has a higher molar specific heat capacity ($6.17 , \text{cal/mol K}$) than the other diatomic gases listed. This is somewhat larger, indicating possible significant contributions from:
Vibrational Modes: In chlorine gas, the vibrational modes might be contributing at room temperature.
Rotational and Translational Modes: The combination of all degrees of freedom.
The higher molar specific heat capacity for chlorine suggests that additional energy levels (such as vibrational) are accessible at room temperature, contributing to the internal energy of the molecules.
Conclusion:
The measured molar specific heats of the listed diatomic gases (H₂, N₂, O₂, NO, CO) being around $5 , \text{cal/mol K}$ are due to their rotational and translational degrees of freedom being accessible at room temperature. Chlorine's slightly higher value indicates vibrational mode contributions. In contrast, monatomic gases, having only translational degrees of freedom, have a lower specific heat capacity of approximately $2.92 , \text{cal/mol K}$.
A child running a temperature of $101^{\circ} \mathrm{F}$ is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to $98^{\circ} \mathrm{F}$ in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is $30 \mathrm{~kg}$. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about $580 \mathrm{cal} \mathrm{g}^{-1}$.
The average rate of extra evaporation caused by the drug is approximately: $$ \text{Average rate of evaporation} \approx 0.0000716 \mathrm{~kg/s} $$ or $$ \text{Average rate of evaporation} \approx 71.6 \mathrm{~mg/s} $$
A 'thermacole' icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side $30 \mathrm{~cm}$ has a thickness of $5.0 \mathrm{~cm}$. If $4.0 \mathrm{~kg}$ of ice is put in the box, estimate the amount of ice remaining after $6 \mathrm{~h}$. The outside temperature is $45^{\circ} \mathrm{C}$, and co-efficient of thermal conductivity of thermacole is $0.01 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$. [Heat of fusion of water $=335 \times 10^{3}$ $\left.\mathrm{J} \mathrm{kg}^{-1}\right]$
Let's break down the solution for your query step by step:
Surface Area Calculation: The surface area of a cube with side (30 , \text{cm}): [ \text{Surface Area} = 6 \times \text{side}^2 = 6 \times (30 , \text{cm})^2 = 5400 , \text{cm}^2 = 0.54 , \text{m}^2 ]
Heat Transfer Calculation: Given:
Thickness of thermacole, $L = 5 \text{cm} = 0.05 \text{m}$
Thermal conductivity, $K = 0.01 \text{J} \text{s}^{-1} \text{m}^{-1} \text{K}^{-1}$
Outside temperature, $T_{outside} = 45^{\circ}C = 318 K$
Inside temperature, $T_{inside} = 0^{\circ}C = 273 K$
Time duration, $(t = 6 \text{hours} = 21600 \text{seconds}$
The rate of heat transfer through the thermacole is given by: [ H = \frac{KA(T_{outside} - T_{inside})}{L} ]
Substituting the values: [ H = \frac{0.01 , \text{J} , \text{s}^{-1} , \text{m}^{-1} , \text{K}^{-1} \times 0.54 , \text{m}^2 \times (318 , \text{K} - 273 , \text{K})}{0.05 , \text{m}} ] [ H = \frac{0.01 \times 0.54 \times 45}{0.05} ] [ H \approx 4.86 , \text{J/s} ]
Total Heat Transfer: The total heat transferred over $6 \text{hours}$: [ Q = H \times t = 4.86 , \text{J/s} \times 21600 , \text{s} = 104976 , \text{J} ]
Ice Melting Calculation: Given:
Heat of fusion of water, $L_f = 335 \times 10^3 , \text{J/kg}$
Mass of ice, $m_{ice} = 4.0 , \text{kg}$
The amount of heat required to melt all the ice: [ Q_{melt} = m_{ice} \times L_f = 4.0 , \text{kg} \times 335 \times 10^3 , \text{J/kg} = 1340000 , \text{J} ]
The fraction of ice melted due to the heat transferred is: [ \text{Fraction} = \frac{Q}{Q_{melt}} = \frac{104976 , \text{J}}{1340000 , \text{J}} \approx 0.0783 ]
Therefore, the mass of ice melted: [ \Delta m = 0.0783 \times 4.0 , \text{kg} \approx 0.313 , \text{kg} ]
So, the amount of ice remaining after $6 \text{hours}$ is: [ 4.0 , \text{kg} - 0.313 , \text{kg} \approx 3.687 , \text{kg} ]
Summary
Approximately 3.69 kg of ice will remain in the icebox after 6 hours.
A brass boiler has a base area of $0.15 \mathrm{~m}^{2}$ and thickness $1.0 \mathrm{~cm}$. It boils water at the rate of $6.0 \mathrm{~kg} / \mathrm{min}$ when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass $=109 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}$ $\mathrm{K}^{-1}$; Heat of vaporisation of water $=2256 \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}$.
The temperature of the part of the flame in contact with the boiler can be found using the solution:
[ T = \frac{15040}{109} ]
Let's compute this:
[ T = \frac{15040}{109} \approx 138.0 \ \mathrm{K} ]
Thus, the temperature of the part of the flame in contact with the boiler is approximately 138.0 K.
Explain why :
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water
(a) A body with large reflectivity is a poor emitter
Reflectivity and Emissivity Relationship: According to Kirchhoff's law of thermal radiation, the emissivity ($ \epsilon$) of a body is equal to its absorptivity ($ \alpha$). This means that a body that is a good reflector (low absorptivity) will be a poor emitter. When a body reflects most of the incident radiation, it absorbs little, and thus emits little radiation as well.
High Reflectivity: A body with large reflectivity ($ R$) has a large proportion of incident light being reflected rather than absorbed. Therefore, its emissivity ($ \epsilon \approx 1 - R $) is low, meaning it emits very little radiation.
(b) A brass tumbler feels much colder than a wooden tray on a chilly day
Thermal Conductivity Difference: Brass has a much higher thermal conductivity than wood. This means that brass can transfer heat away from your hand much more efficiently than wood can.
Heat Transfer: When you touch the brass tumbler, it quickly draws heat from your hand, making it feel colder. In contrast, the wooden tray, with its lower thermal conductivity, does not draw heat away from your hand as quickly, and thus feels warmer.
(c) An optical pyrometer gives too low a value for the temperature of a red hot iron piece in the open but the correct value in the furnace
Emissivity Differences: An optical pyrometer is calibrated for ideal black body radiation, which assumes an emissivity of 1. A red hot iron piece in the open air will have a lower emissivity compared to when it is in a furnace.
Black Body Environment: In a furnace, the walls are also hot, and multiple reflections ensure that the iron piece behaves more like a black body (higher effective emissivity). In open air, with fewer reflections and lower surrounding temperature, the emissivity is lower, leading to an underestimation of temperature by the pyrometer calibrated for a black body.
(d) The earth without its atmosphere would be inhospitably cold
Greenhouse Effect: The Earth's atmosphere contains gases like carbon dioxide and water vapor that trap heat. This is known as the greenhouse effect.
Heat Retention: These gases allow sunlight to enter but inhibit the longer-wavelength infrared radiation from escaping, maintaining the Earth's temperature within a habitable range.
Absence of Atmosphere: Without the atmosphere, much of the heat would be lost to space, leading to a much colder surface temperature, rendering the Earth inhospitably cold.
(e) Heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water
Latent Heat Transfer: Steam heating systems use the latent heat of vaporization. Steam releases a large amount of heat when it condenses back into water.
Higher Heat Content: The phase change from steam to water transfers more energy compared to raising the temperature of water by the same amount.
Efficient Heat Distribution: Steam can travel faster through pipes and can reach higher temperatures, providing more uniform and efficient heating compared to hot water systems.
By considering these principles, you can understand the specific properties and behaviors related to thermal conductivity, radiation, emissivity, and the phase changes of water, which underline the efficiency and effectiveness of different heating and temperature measurement systems.
A body cools from $80^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 5 minutes. Calculate the time it takes to cool from $60^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$. The temperature of the surroundings is $20^{\circ} \mathrm{C}$.
We've calculated the required values as follows:
$ k = 0.13333 $
To find the time for the final cooling scenario:
The time to cool from $60^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$: $$ \text{Time} = \frac{6}{5k} = \frac{6}{5 \times 0.13333} \approx 9.0 \text{ minutes} $$
Therefore, it takes approximately 9.0 minutes to cool from $60^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$.
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Notes - Thermal Properties of Matter | Class 11 NCERT | Physics
Comprehensive Guide to Thermal Properties of Matter: Class 11 Notes
Introduction
Thermal properties of matter are crucial for understanding how substances respond to changes in temperature and heat. For Class 11 students, grasping these concepts is essential for advancing in physics.
Basics of Temperature and Heat
What is Temperature?
Temperature is a measurable indication of the hotness or coldness of a body. It is a relative measure, meaning that comparisons tell us which object is hotter or colder. For example, boiling water is hotter than an ice cube. Unlike temperature, terms like "hot" and "cold" are subjective and vary based on the observer's perception.
Understanding Heat
Heat is a form of energy transferred between two systems or between a system and its surroundings due to a temperature difference. Heat flows from a hotter body to a colder one until thermal equilibrium is reached. This transfer can cause physical changes such as an increase in temperature, expansion, or change in state (e.g., melting, boiling).
Measurement of Temperature
Thermometers and Their Types
Temperature measurement is usually done with a thermometer. The most commonly used type is the liquid-in-glass thermometer, which uses mercury or alcohol. These liquids expand and contract uniformly with temperature changes, giving a precise measurement.
Temperature Scales
Celsius: Defined by the freezing point (0°C) and boiling point (100°C) of water. Fahrenheit: Defines the freezing point of water at 32°F and the boiling point at 212°F. Kelvin: The absolute temperature scale where 0 K is absolute zero.
To convert between Celsius and Fahrenheit: [ t_F = \left(\frac{9}{5}\right) t_C + 32 ]
Heat Transfer Mechanisms
Conduction
Conduction is the transfer of heat through direct contact. It occurs in solids where atoms vibrate and pass on energy to neighbouring atoms. Metals are excellent conductors of heat due to their free electrons.
Convection
Convection occurs in fluids (liquids and gases) where the warmer part of the fluid rises and the cooler part sinks, creating a circulation pattern. This movement transfers heat throughout the fluid.
Radiation
Radiation is the transfer of heat through electromagnetic waves. It does not require any medium, hence why we feel the warmth of the sun even across the vacuum of space.
Ideal-Gas Equation and Absolute Temperature
What is an Ideal Gas?
An ideal gas is a theoretical gas composed of particles that do not interact except for elastic collisions. The behaviour of gases can be described by the ideal-gas equation: [ PV = \mu RT ] where (P) represents pressure, (V) volume, (T) temperature, (\mu) the number of moles, and (R) the universal gas constant.
Thermal Expansion
Types of Thermal Expansion
- Linear Expansion: Change in length (\frac{\Delta l}{l} = \alpha_l \Delta T )
- Area Expansion: Change in area
- Volume Expansion: Change in volume (\frac{\Delta V}{V} = \alpha_v \Delta T )
Different materials expand at different rates due to their unique coefficients of expansion. This is why metals generally expand more than glass or ceramics.
Specific Heat Capacity and Calorimetry
What is Specific Heat Capacity?
Specific heat capacity is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius. [ s = \frac{1}{m} \frac{\Delta Q}{\Delta T} ] where (s) is specific heat capacity, (m) is mass, (\Delta Q) is heat added, and (\Delta T) is temperature change.
Calorimetry
Calorimetry is the measurement of heat transfer from one body to another. It involves using a calorimeter to determine heat capacity and other thermal properties.
Change of State and Latent Heat
State Changes
Matter can change from one state to another when heated or cooled. Common state changes include:
- Melting: Solid to liquid
- Freezing: Liquid to solid
- Vaporisation: Liquid to gas
- Sublimation: Solid to gas
Latent Heat
Latent heat is the heat required for a substance to change state without a change in temperature. It includes the latent heat of fusion (melting/freezing) and the latent heat of vaporisation (boiling/condensation).
Heat Transfer Processes
Conduction
Quantitatively described by the equation: [ H = KA \frac{T_C - T_D}{L} ] where (H) is heat transfer rate, (K) is thermal conductivity, (A) is the cross-sectional area, (T_C) and (T_D) are temperatures, and (L) is the length.
Convection
Involves bulk movement of fluid masses. For example, sea breezes occur due to differential heating of land and water.
Radiation
Described by Stefan-Boltzmann law: [ H = A \sigma T^4 ] where (H) is the radiated heat, (A) is the area, (\sigma) is the Stefan-Boltzmann constant, and (T) is temperature.
Newton’s Law of Cooling
This law states that the rate of cooling is proportional to the temperature difference between an object and its surroundings: [ -\frac{dQ}{dt} = k(T_2 - T_1) ] This equation helps us understand cooling processes and estimate cooling times.
Conclusion
Understanding thermal properties of matter is fundamental for multiple scientific and practical applications. These concepts help explain everyday phenomena and are critical for advanced studies in physics. Class 11 students should thoroughly explore these principles to build a strong foundation in thermal physics.
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