Heat and Thermodynamics

The efficiency of a Carnot engine is calculated using the formula:

$ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} $

Initially, the efficiency $\eta_1$ is:

$ \eta_1 = 1 - \frac{2}{3} = \frac{1}{3} $

After the change in the temperature ratio, the efficiency $\eta_2$ becomes:

$ \eta_2 = 1 - \frac{3}{4} = \frac{1}{4} $

To determine the change in efficiency, we calculate $\Delta \eta$:

$ \Delta \eta = \eta_2 - \eta_1 = \frac{1}{4} - \frac{1}{3} = -\frac{1}{12} $

The percentage change in efficiency is calculated as follows:

$ \frac{\Delta \eta}{\eta_1} \times 100\% = \frac{-\frac{1}{12}}{\frac{1}{3}} \times 100 = -\frac{1}{4} \times 100\% = -25\% $

The negative sign indicates that there is a 25% decrease in efficiency.

The molar specific heat capacities of gases at constant pressure vary depending on whether the gas is monoatomic or diatomic due to differences in their degrees of freedom.

For a monoatomic gas, the molar specific heat capacity at constant pressure, denoted as $ C_{p_m} $, is given by:

$ C_{p_m} = \frac{5}{2} R $

For a diatomic gas, the molar specific heat capacity at constant pressure, denoted as $ C_{p_d} $, is:

$ C_{p_d} = \frac{7}{2} R $

The ratio of the molar specific heat capacities for monoatomic and diatomic gases is therefore:

$ \frac{C_{p_m}}{C_{p_d}} = \frac{\left(\frac{5}{2} R\right)}{\left(\frac{7}{2} R\right)} = \frac{5 \times 2 \times R}{2 \times 7 \times R} = \frac{5}{7} $

Thus, the ratio is $ 5:7 $.

To find the mass of water $ m $ that results in a final mixture temperature of $ 6^\circ \mathrm{C} $, let's break down the heat exchanges that occur during the process.

Step 1: Heating Ice from $-20^\circ \mathrm{C}$ to $0^\circ \mathrm{C}$

The heat required is given by:

$ Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T = 5\, \text{g} \times 0.5\, \text{cal/g}{}^\circ\mathrm{C} \times 20^\circ\mathrm{C} = 50\, \text{cal} $

Step 2: Melting Ice at $0^\circ \mathrm{C}$ to Water

The heat required for melting the ice is:

$ Q_2 = m_{\text{ice}} \times L_f = 5\, \text{g} \times 80\, \text{cal/g} = 400\, \text{cal} $

Step 3: Heating Water from Ice from $0^\circ \mathrm{C}$ to $6^\circ \mathrm{C}$

The heat required to raise the temperature of the resulting water is:

$ Q_3 = m_{\text{water from ice}} \cdot c_{\text{water}} \cdot \Delta T = 5\, \text{g} \times 1\, \text{cal/g}{}^\circ\mathrm{C} \times 6^\circ\mathrm{C} = 30\, \text{cal} $

Total Heat Required

Summing these components, the total heat absorbed by the ice and converted water is:

$ Q_{\text{total gained}} = Q_1 + Q_2 + Q_3 = 50 + 400 + 30 = 480\, \text{cal} $

Step 4: Heat Lost by Water Initially at $30^\circ \mathrm{C}$ Cooling to $6^\circ \mathrm{C}$

The heat lost is calculated by:

$ Q_{\text{lost}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T = m \times 1\, \text{cal/g}{}^\circ\mathrm{C} \times 24^\circ\mathrm{C} = 24m\, \text{cal} $

Equilibrium Condition

The heat gained must equal the heat lost:

$ Q_{\text{total gained}} = Q_{\text{lost}} $

$ 480 = 24m $

Solving for $ m $:

$ m = \frac{480}{24} = 20\, \text{g} $

Therefore, the mass of water required is $ 20\, \text{g} $.

Work Done During Isothermal Expansion:

For an ideal gas expanding isothermally, the work done $ W $ is given by:

$ W = nRT \ln\left(\frac{V_f}{V_i}\right) $

where $ n $ is the number of moles, $ R $ is the ideal gas constant, $ T $ is the temperature, $ V_f $ is the final volume, and $ V_i $ is the initial volume.

Gas A Expansion:

The pressure of gas $ A $ decreases by 50%, thus:

$ p_f = 0.5 p_i $

Using the ideal gas law $ pV = nRT $, we have:

$ p_f V_f = p_i V_i $

$ 0.5 p_i V_f = p_i V_i \Rightarrow V_f = 2 V_i, \text{ thus } \frac{V_f}{V_i} = 2 $

Therefore, the work done by gas $ A $ is:

$ W_A = nRT_1 \ln(2) $

Gas B Expansion:

The pressure of gas $ B $ decreases by 75%, thus:

$ p_f = 0.25 p_i $

Using the ideal gas law again:

$ p_f V_f = p_i V_i $

$ 0.25 p_i V_f = p_i V_i \Rightarrow V_f = 4 V_i, \text{ thus } \frac{V_f}{V_i} = 4 $

The work done by gas $ B $ is:

$ W_B = nRT_2 \ln(4) $

Determining the Temperature Ratio

Given that the work done by both gases is the same, $ W_A = W_B $, we equate the expressions for work:

$ nRT_1 \ln(2) = nRT_2 \ln(4) $

Solving for the temperature ratio:

$ T_1 \ln(2) = T_2 \ln(4) $

$ T_1 \ln(2) = T_2 \cdot 2 \ln(2) $

$ \frac{T_1}{T_2} = \frac{2 \ln(2)}{\ln(2)} = 2 $

Thus, the ratio of temperatures $ T_1 : T_2 $ is:

$ 2:1 $

Given:

Heat absorbed, $ Q = 80 \, \text{J} $

Change in volume, $ \Delta V = 16 \times 10^{-5} \, \text{m}^3 $

First Law of Thermodynamics

$ Q = \Delta U + W $

For a monoatomic ideal gas, the change in internal energy ($ \Delta U $) can be expressed as:

$ \Delta U = \frac{3}{2} n R \Delta T $

The work done ($ W $) by the gas during expansion is:

$ W = p \Delta V $

Using the ideal gas law:

$ \begin{aligned} & p V = n R T \\ & p \Delta V = n R \Delta T \end{aligned} $

Substitute into the internal energy equation:

$ \Delta U = \frac{3}{2} p \Delta V $

Now substitute equations for $ W $ and $ \Delta U $ into the first law:

$ \begin{aligned} Q & = \frac{3}{2} p \Delta V + p \Delta V \\ 80 & = \frac{5}{2} \times p \times 16 \times 10^{-5} \end{aligned} $

Solve for pressure $ p $:

$ \begin{aligned} \frac{2 \times 80}{16 \times 10^{-5} \times 5} & = p \\ \frac{160}{80 \times 10^{-5}} & = p \\ 2 \times 10^5 \, \text{N/m}^2 & = p \end{aligned} $

Thus, the pressure of the gas is $ 2 \times 10^5 \, \text{N/m}^2 $.

The efficiency $ (\eta) $ of a Carnot engine is determined by the formula:

$ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} $

Given that the initial efficiency is $ 25\% $ and the source temperature is $ 127^\circ \mathrm{C} $, which converts to $ 400 \, \mathrm{K} $ in absolute temperature, we have:

$ 0.25 = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} $

This implies:

$ \frac{T_{\text{sink}}}{T_{\text{source}}} = 0.75 $

From which:

$ T_{\text{sink}} = 0.75 \times 400 = 300 \, \mathrm{K} $

If we decrease the sink temperature by $ 10\% $, the new sink temperature $ T_{\text{sink}}^{\prime} $ becomes:

$ T_{\text{sink}}^{\prime} = T_{\text{sink}} - \frac{T_{\text{sink}} \times 10}{100} = 300 - 30 = 270 \, \mathrm{K} $

The new efficiency $ \eta^{\prime} $ of the engine is calculated as follows:

$ \eta^{\prime} = 1 - \frac{T_{\text{sink}}^{\prime}}{T_{\text{source}}} \times 100\% $

Substituting the values, we get:

$ \eta^{\prime} = 1 - \frac{270}{400} \times 100\% $

$ = 1 - 0.675 \times 100\% $

$ \eta^{\prime} = 0.325 \times 100\% $

Therefore, the new efficiency is:

$ \eta^{\prime} = 32.5\% $

For a monoatomic gas:

Number of moles, $ n_m = 2 $.

Temperature, $\Delta T_m = 27^{\circ} \mathrm{C} = 300\, \mathrm{K}$.

Internal energy, $ U_m = U $.

The total internal energy for a monoatomic gas is calculated as:

$ U_m = \frac{3}{2} n R \Delta T $

Substituting the values:

$ U = \frac{3}{2} \times 2 \times R \times 300 $

So,

$ U = 900R \quad \text{...(i)} $

For a diatomic gas:

Number of moles, $ n_d = 3 $.

Temperature, $\Delta T_d = 127^{\circ} \mathrm{C} = 400\, \mathrm{K}$.

The total internal energy for a diatomic gas is given by:

$ U_d = \frac{5}{2} n R \Delta T $

Substitute the known values:

$ U^{\prime} = \frac{5}{2} \times 3 \times R \times 400 $

Thus,

$ U^{\prime} = 3000R \quad \text{...(ii)} $

To find the relationship between $ U^{\prime} $ and $ U $, divide Eq. (ii) by Eq. (i):

$ \frac{U^{\prime}}{U} = \frac{3000R}{900R} $

Therefore,

$ U^{\prime} = \frac{10}{3} U $

The specific details are as follows:

Mass of water, $m_w = 200 \, \text{g}$

Specific heat capacity of water, $C_w = 1 \, \text{cal/g}^\circ \text{C}$

Change in temperature of the water, $\Delta T_w = 10^\circ \text{C}$ (from $80^\circ \text{C}$ to $70^\circ \text{C}$)

Mass of the metal, $m_m = 100 \, \text{g}$

Change in temperature of the metal, $\Delta T_m = 50^\circ \text{C}$ (from $20^\circ \text{C}$ to $70^\circ \text{C}$)

Calculating Heat Exchanged

Using the heat equation, we calculate:

Heat lost by the water:

$ Q_w = m_w \times C_w \times \Delta T_w $

Heat gained by the metal:

$ Q_m = m_m \times C_m \times \Delta T_m $

Setting Heat Lost Equal to Heat Gained

The principle of conservation of energy tells us that the heat lost by the water equals the heat gained by the metal:

$ m_w \times C_w \times \Delta T_w = m_m \times C_m \times \Delta T_m $

Substituting the values, we have:

$ 200 \times 1 \times 10 = 100 \times C_m \times 50 $

By solving for $C_m$:

$ C_m = \frac{200 \times 1 \times 10}{100 \times 50} = \frac{2}{5} \, \text{cal/g}^\circ \text{C} $

Calculating the Ratio

Finally, the ratio of the specific heat capacity of the metal to that of water is:

$ \frac{C_m}{C_w} = \frac{\frac{2}{5}}{1} = \frac{2}{5} $

To determine the efficiency of a heat engine operating between two specific temperatures, we use the formula:

$ \eta = 1 - \frac{T_C}{T_H} $

where:

$ T_C $ is the temperature of the cold reservoir.

$ T_H $ is the temperature of the hot reservoir.

From the temperature scale relationships:

The Celsius and Fahrenheit scales coincide at $-40^\circ$, which is equivalent to 233.15 Kelvin.

The Kelvin and Fahrenheit scales coincide at 574.59 Kelvin.

Using these temperatures, we calculate the efficiency:

$ \begin{aligned} \eta &= 1 - \frac{233.15}{574.59} \\ &\approx 1 - 0.4056 \\ &\approx 0.5944 \end{aligned} $

Therefore, the efficiency of the heat engine operating between these temperatures is approximately 59.44%.

Given the initial conditions of the problem:

Pressure, $ p_1 = 10^5 \, \mathrm{Nm}^{-2} $

Initial volume, $ V_1 = 16 \, \mathrm{L} $

Final volume, $ V_2 = 2 \, \mathrm{L} $

The molar specific heat at constant volume is given as $ \frac{3}{2} R $.

For an adiabatic process, the work done on the gas is calculated using the formula:

$ W = \frac{n R \Delta T}{\gamma-1} = \frac{p_2 V_2 - p_1 V_1}{\gamma - 1} $

First, we determine the final pressure $ p_2 $ using the adiabatic relation:

$ p_2 = p_1 \left(\frac{V_1}{V_2}\right)^\gamma $

Substituting the given values:

$ p_2 = 10^5 \left(\frac{16}{2}\right)^{1.66} = 3.15 \times 10^6 \, \mathrm{N/m}^2 $

Next, we calculate the work done:

$ W = \frac{\left(3.15 \times 10^6 \times 2\right) - \left(10^5 \times 16\right)}{1.66 - 1} $

Simplifying further:

$ W = \frac{6.3 \times 10^6 - 16 \times 10^5}{0.66} $

$ W = \frac{63 \times 10^5 - 16 \times 10^5}{0.66} $

$ W = \frac{47 \times 10^5}{0.66} = 71.21 \times 10^5 $

Finally, converting the work done to kilojoules:

$ \text{Work done} = 71.21 \times 10^5 \times 10^{-3} \, \mathrm{J} = 7.1 \, \mathrm{kJ} $

Thus, the work done on the gas is 7.1 kJ.

The work done during a cycle is equal to the area under $p$ - $V$ diagram.

$ \begin{aligned} &\text { From the given } p \text { - } V \text { diagram, }\\ &\begin{aligned} W & =\frac{1}{2} \times A C \times B C \\ W & =\frac{1}{2} \times(3 V-V) \times(3 p-p) \\ & =\frac{1}{2} \times 2 V \times 2 p \\ W & =2 p V \end{aligned} \end{aligned} $

At NTP, the internal energy of a diatomic gas is primarily due to translational and rotational motions, which is given by $\frac{5}{2} k_B T$.

At high, temperature, vibrational modes becomes accessible and internal energy increases because these modes contribute additional degree of freedom. The internal energy increases because these modes contribute in temperature, $\frac{7}{2} k_B T$.

Therefore, the ratio of KE,

$\frac{\text { KE at high temperature }}{\text { KE at NTP }}=\frac{7 / 2 k_B T_{\text {high }}}{5 / 2 k_B T_{\text {NTP }}}$

This is a simplified model and assumes that all vibrational modes are fully excited, which may not be the case depending on the exact temperature. All options are incorrect.

Given a metal block composed of various metals:

Aluminium mass: $ m_{\text{Al}} = 2.4 \, \text{kg} $

Brass mass: $ m_{\text{Brass}} = 1.6 \, \text{kg} $

Copper mass: $ m_{\text{Cu}} = 0.8 \, \text{kg} $

The specific heats of the components are:

Aluminium: $ C_{\text{Al}} = 0.216 \, \text{cal} \, \text{kg}^{-1} \, {}^{\circ}\text{C}^{-1} $

Brass: $ C_{\text{Brass}} = 0.0917 \, \text{cal} \, \text{kg}^{-1} \, {}^{\circ}\text{C}^{-1} $

Copper: $ C_{\text{Cu}} = 0.0931 \, \text{cal} \, \text{kg}^{-1} \, {}^{\circ}\text{C}^{-1} $

The heat supplied to the block is 44.4 cal. To find the final temperature of the block, we use the formula for heat energy:

$ Q = m C \Delta T $

The total heat equation is:

$ Q = (m_{\text{Al}} C_{\text{Al}} + m_{\text{Brass}} C_{\text{Brass}} + m_{\text{Cu}} C_{\text{Cu}}) \Delta T $

Substitute the given values:

$ 44.4 = (2.4 \times 0.216 + 1.6 \times 0.0917 + 0.8 \times 0.0931) \Delta T $

Calculate the product inside the brackets:

$ 44.4 = [0.5184 + 0.14672 + 0.07448] \Delta T $

Simplifying further:

$ 44.4 = 0.7396 \Delta T $

Solving for $ \Delta T $:

$ \Delta T = \frac{44.4}{0.7396} $

$ \Delta T = 60 $

The initial temperature $ T_i = 20^{\circ} \text{C} $. Therefore, the final temperature $ T_f $ is:

$ T_f = \Delta T + T_i $

$ T_f = 60 + 20 $

$ T_f = 80^{\circ} \text{C} $

During an adiabatic process, the given relationship is $ p V^{\frac{3}{2}} = \text{constant} $. In general, for an adiabatic process, the equation is $ p V^\gamma = \text{constant} $. By comparing these equations, we deduce that $ \gamma = \frac{3}{2} $.

Consider the initial volume $ V_1 = V_0 $ and the final volume $ V_2 = \frac{V_0}{4} $.

The relation between temperature and volume in an adiabatic process is given by:

$ T V^{\gamma-1} = \text{constant} $

Thus:

$ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} $

Substituting the known values, we have:

$ T \left(V_0\right)^{\gamma-1} = T_2 \left[\frac{V_0}{4}\right]^{\gamma-1} $

This simplifies to:

$ T_2 = T (4)^{\frac{3}{2}-1} $

Therefore, the final temperature is:

$ T_2 = T \sqrt{2} $

The condition $ dW = dQ $ is valid in an isothermal process.

According to the first law of thermodynamics:

$ dQ = dW + dU $

In an isothermal process, the temperature remains constant, which implies $ dU = 0 $. Therefore:

$ \therefore \quad dQ = dW $

To find the temperature of the sink ($T_2$), we begin by analyzing the efficiency increase of a Carnot engine.

Initial Conditions

Initial efficiency: $\eta = 25\%$ or 0.25.

Efficiency formula:

$ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} $

Substitute the given efficiency:

$ 0.25 = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \quad \Rightarrow \quad \frac{T_{\text{sink}}}{T_{\text{source}}} = 0.75 $

Thus, the temperature of the sink is:

$ T_{\text{sink}} = 0.75 \times T_{\text{source}} $

After Increasing the Source Temperature

The source temperature is increased by 100 K:

$ T_{\text{source}}^{\prime} = T_{\text{source}} + 100 $

New efficiency: $\eta^{\prime} = 40\%$ or 0.40. Using the efficiency formula again:

$ \eta^{\prime} = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}^{\prime}} $

Substitute the known values:

$ 0.40 = 1 - \frac{0.75 \times T_{\text{source}}}{T_{\text{source}} + 100} $

Solving for $T_{\text{source}}$, we find:

$ T_{\text{source}} = 400 \text{ K} $

Calculating the Sink Temperature

Substitute $T_{\text{source}}$ in the sink temperature equation:

$ T_{\text{sink}} = 0.75 \times 400 = 300 \text{ K} $

Thus, the temperature of the sink ($T_2$) is 300 K.

For a deeper understanding of how the heat capacity ratio $\gamma = \frac{C_p}{C_v}$ varies with different types of gases based on their degrees of freedom, consider the following explanations:

Monoatomic Gases: The ratio $\gamma$ is typically $\frac{5}{3}$. These gases have three translational degrees of freedom, which leads to this specific ratio.

Diatomic Gases (Rigid): For diatomic gases that behave rigidly (i.e., where vibrational modes are not considered), $\gamma$ is $\frac{7}{5}$. The degrees of freedom include three translational and two rotational.

Diatomic Gases (Non-Rigid): For non-rigid diatomic gases, where vibrational modes are considered, $\gamma$ is $\frac{9}{7}$. Additional vibrational degrees of freedom affect this ratio.

Polyatomic Gases: These gases can have more complex molecular structures, resulting in more degrees of freedom. Here, for any polyatomic gas, the heat capacities are given by $C_p = 4 + f$ and $C_V = 3 + f$, where $f$ represents the additional degrees of freedom available to the gas. Therefore, the ratio $\gamma$ is expressed as:

$ \gamma = \frac{C_p}{C_V} = \frac{4 + f}{3 + f} $

Using this understanding, the correct pairing is as follows:

A (Monoatomic): Corresponds to II ($\frac{5}{3}$)

B (Diatomic Rigid): Corresponds to III ($\frac{7}{5}$)

C (Diatomic Non-Rigid): Corresponds to IV ($\frac{9}{7}$)

D (Polyatomic): Corresponds to I ($\frac{4+f}{3+f}$)

Consequently, the correct match is A-II, B-III, C-IV, D-I.

Given,

Area of thickness is same for both plates.

Free face of brass $=0^{\circ} \mathrm{C}$

Free face of copper $=100^{\circ} \mathrm{C}$

$ \frac{K_B}{K_C}=\frac{1}{4} \Rightarrow K_C=4 K_B $

$ \begin{aligned} &\text { Let, the temperature of interface }=T^{\circ} \mathrm{C}\\ &100^{\circ} \mathrm{C} \quad T^{\circ} \mathrm{C} \quad 0^{\circ} \mathrm{C} \end{aligned} $

$ \begin{array}{|l|l|} \hline \text { Copper } & \text { Brass } \\ \hline \end{array} $

Heat transfer by copper = Heat gained by brass

$ \begin{aligned} & \left(\frac{Q}{t}\right)_{\mathrm{Cu}}=\left(\frac{Q}{t}\right)_{\text {brass }} \\ & \Rightarrow \quad \frac{K_C A\left(100^{\circ}-T\right)}{l}=\frac{K_B A\left(T-0^{\circ}\right)}{l} \\ & \Rightarrow \quad 4 K_B\left(100^{\circ}-T\right)=K_B(T-0) \\ & \Rightarrow \quad 400-4 T=T \Rightarrow 400=5 T \\ & \Rightarrow \quad T=\frac{400}{5}=80^{\circ} \mathrm{C} \end{aligned} $

For an ideal monatomic gas, the internal energy depends only on the temperature, not on the process by which the temperature change occurs. Here’s a quick breakdown:

The internal energy of a monatomic ideal gas is given by:

$U = \frac{3}{2} nRT$

where:

$n$ is the number of moles,

$R$ is the gas constant,

$T$ is the temperature.

The change in internal energy when the gas is heated from $T_1$ to $T_2$ is therefore:

$\Delta U = \frac{3}{2} nR (T_2 - T_1)$

This expression is valid regardless of whether the heating occurs at constant volume or constant pressure, because the internal energy is a state function—it depends only on the initial and final temperatures.

Thus, the change in internal energy is the same in both cases.

Correct Answer: Option C.

When the temperatures are $ T_1 = 200^{\circ} \mathrm{C} = 473 \mathrm{~K} $ and $ T_2 = 0^{\circ} \mathrm{C} = 273 \mathrm{~K} $, the efficiency $\eta_1$ of the Carnot engine is calculated as:

$ \eta_1 = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{473} = 1 - 0.577 = 0.423 $

When the temperatures are $ T_1 = 0^{\circ} \mathrm{C} = 273 \mathrm{~K} $ and $ T_2 = -200^{\circ} \mathrm{C} = 73 \mathrm{~K} $, the efficiency $\eta_2$ is calculated as:

$ \eta_2 = 1 - \frac{T_2}{T_1} = 1 - \frac{73}{273} = 1 - 0.267 = 0.733 $

Therefore, the ratio of the efficiencies $\frac{\eta_1}{\eta_2}$ is given by:

$ \frac{\eta_1}{\eta_2} = \frac{0.423}{0.733} \approx 0.577 $

Thus, $\frac{\eta_1}{\eta_2} \approx 0.58$.

We know that,

In a cyclic process, $\Delta U=0$

According to first law of thermodynamics,

$ \begin{aligned} & \Delta Q=\Delta U+W \\ & \Delta Q=0+\text { Area under } p-V \text { curve } \end{aligned} $

$ \begin{aligned} & \Delta Q=\pi \times\left(10 \times 10^3\right) \times\left(10 \times 10^{-3}\right) \\ & \Delta Q=\pi \times 10 \times 10 \\ & \Delta Q=10^2 \pi \mathrm{~J} \end{aligned} $

According to the law of equipartition of energy, the energy associated with each degree of freedom of a molecule is $\frac{1}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the temperature in Kelvin.

For a polyatomic gas with $n$ degrees of freedom, the total mean kinetic energy per molecule can be calculated as:

$ \text{Mean kinetic energy per molecule} = \frac{1}{2} n k_B T $

This formula reflects that each degree of freedom contributes $\frac{1}{2} k_B T$ to the total energy, and the number of degrees of freedom, $n$, multiplies this contribution to give the total mean kinetic energy of each molecule in the gas.

A perfect black body is defined as an object that absorbs all incident electromagnetic radiation. This means the fraction of radiation it absorbs, known as the absorption coefficient or absorptivity, is exactly $1$.

Here's a brief explanation:

The absorption coefficient represents the proportion of radiation absorbed.

For a perfect black body, by definition, no radiation is reflected or transmitted.

Therefore, the absorption coefficient is $1$, meaning it absorbs 100% of the incident radiation.

So, the correct answer is:

Option D: $1$.

Given that the initial temperature $ T_1 = 300 \, \text{K} $, the ratio of specific heats $ \gamma = 1.50 $, and the final volume $ V_2 = 2V_1 $, we can calculate the resultant fall in temperature during adiabatic expansion using the following process:

For an adiabatic expansion, the relation is given by:

$ TV^{\gamma-1} = \text{constant} $

This implies:

$ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} $

Substitute the given values:

$ T_2 = T_1 \left(\frac{1}{2}\right)^{1.50-1} = 300 \left(\frac{1}{2}\right)^{0.5} $

Calculating $ T_2 $, we find:

$ T_2 = 213.13 \, \text{K} $

Thus, the resultant fall in temperature is:

$ = T_1 - T_2 \approx 88 \, \text{K} $

To solve this problem, we need to find how much we must increase the temperature of the source to double the efficiency of the Carnot engine from 25% to 50%.

Given:

Sink temperature, $ T_L = 300 \, \text{K} $

Initial efficiency, $ \eta = 25\% = 0.25 $

Step 1: Calculate the Original Source Temperature

Using the efficiency formula for a Carnot engine:

$ \eta = 1 - \frac{T_L}{T_H} $

Substitute the known values:

$ 0.25 = 1 - \frac{300}{T_H} $

Solving for $ T_H $:

$ T_H = \frac{300}{1 - 0.25} = \frac{300}{0.75} = 400 \, \text{K} \quad \ldots \text{ (i) } $

Step 2: Calculate the New Source Temperature for $ 50\% $ Efficiency

New efficiency, $ \eta^{\prime} = 50\% = 0.50 $

Using the efficiency formula again:

$ \eta^{\prime} = 1 - \frac{T_L}{T_H^{\prime}} $

Substitute the values:

$ 0.50 = 1 - \frac{300}{T_H^{\prime}} $

Solving for $ T_H^{\prime} $:

$ T_H^{\prime} = \frac{300}{1 - 0.50} = \frac{300}{0.50} = 600 \, \text{K} \quad \ldots \text{ (ii) } $

Step 3: Calculate the Increase in Temperature

The increase in temperature required is:

$ = T_H^{\prime} - T_H = 600 \, \text{K} - 400 \, \text{K} = 200 \, \text{K} $

Thus, the required increase in the source temperature is 200 K.

Given the following:

Heat supplied, $ Q = 100\, \text{J} $

Increase in internal energy, $ \Delta U = 60\, \text{J} $

Using the first law of thermodynamics, we have:

$ Q = W + \Delta U $

Substitute the given values:

$ 100 = W + 60 $

$ W = 40\, \text{J} \quad \ldots \text{(i)} $

For a monoatomic gas, the heat capacity ratio $ \gamma = \frac{5}{3} $.

The ratio of work done ($ \Delta W $) to the heat supplied ($ \Delta Q $) for a monoatomic gas is:

$ \frac{\Delta W}{\Delta Q} = \frac{\gamma - 1}{\gamma} = \frac{\frac{5}{3} - 1}{\frac{5}{3}} = 0.4 \text{ or } 40\% \quad \ldots \text{(ii)} $

This calculation indicates that 40% of the supplied heat is converted into work done in a monoatomic gas.

Equation (i) confirms this analysis.

Therefore, the gas is monoatomic in nature.

Given:

Volume, $ V = 3 \, \text{m}^3 $

Pressure, $ p = 3 \times 10^5 \, \text{Pa} $

The internal energy of an ideal gas can be calculated using the formula:

$ E = \frac{3}{2} p V $

Substituting the given values:

$ E = \frac{3}{2} \times 3 \times 10^5 \times 3 $

Calculating the above expression gives:

$ E = 13.5 \times 10^5 \, \text{J} $

To determine the ratio of the lengths of two cylindrical rods, let's review the thermal conductivity formula:

$ K = \frac{Q \cdot L}{A \cdot \Delta T} $

Where:

$ Q $ is the rate of heat flow.

$ L $ is the length of the rod.

$ A $ is the cross-sectional area.

$ \Delta T $ is the temperature difference across the rod.

Since the rods $ A $ and $ B $ are made of the same material, their thermal conductivity, $ K $, is equal. This gives us:

$ K_A = \frac{Q_A \cdot L_A}{A_A \cdot \Delta T_A} \quad \text{and} \quad K_B = \frac{Q_B \cdot L_B}{A_B \cdot \Delta T_B} $

Given that the rods have the same mass and material, their volumes are equal:

$ V_A = V_B $

Thus, the equation for equal volumes becomes:

$ A_A \cdot L_A = A_B \cdot L_B $

From this, we derive:

$ \frac{L_A}{L_B} = \frac{A_B}{A_A} $

Since $ K_A = K_B $, we equate the expressions:

$ \frac{Q_A \cdot L_A}{A_A \cdot \Delta T_A} = \frac{Q_B \cdot L_B}{A_B \cdot \Delta T_B} $

Substituting the ratio of volumes, we get:

$ \frac{L_A}{L_B} = \frac{Q_B}{Q_A} \times \frac{A_A}{A_B} \times \frac{\Delta T_A}{\Delta T_B} $

Using the given values:

$ \frac{L_A}{L_B} = \frac{8}{3} \times \frac{L_B}{L_A} \times \frac{40}{60} $

Solving the equation:

$ \frac{L_A^2}{L_B^2} = \frac{8}{3} \times \frac{4}{6} = \frac{32}{18} = \frac{16}{9} $

Taking the square root gives:

$ \frac{L_A}{L_B} = \sqrt{\frac{16}{9}} = \frac{4}{3} $

Thus, the ratio of the lengths of rods $ A $ and $ B $ is $ \frac{4}{3} $.

Given the efficiency of the Carnot cycle, $\eta = \frac{1}{6}$, and a scenario where the sink's temperature is lowered by 65 K, resulting in an increased efficiency of $\eta_2 = \frac{1}{3}$.

Efficiency of the Carnot Cycle

The efficiency $\eta$ of a Carnot cycle is expressed by:

$ \eta_1 = 1 - \frac{T_2}{T_1} $

where $T_2$ is the temperature of the sink, and $T_1$ is the temperature of the source.

Given:

$ \frac{1}{6} = 1 - \frac{T_2}{T_1} $

Now, when the sink's temperature is reduced by 65 K, the efficiency becomes:

$ \eta_2 = 1 - \frac{(T_2 - 65)}{T_1} $

Calculation

Substitute the expression for $\frac{T_2}{T_1}$ from the first situation into the new efficiency equation:

$ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} $

So, for the new condition:

$ \frac{1}{3} = 1 - \left( \frac{T_2}{T_1} - \frac{65}{T_1} \right) $

Substitute the value from the previous step:

$ \frac{65}{T_1} = \frac{1}{3} - \frac{1}{6} = \frac{3}{18} = \frac{1}{6} $

Solve for $T_1$:

$ T_1 = 65 \times 6 = 390 \, \text{K} $

Initial Temperature of Sink

Calculate $T_2$ using the ratio determined earlier:

$ \frac{T_2}{T_1} = \frac{5}{6} $

Thus:

$ T_2 = \frac{5}{6} \times 390 = 325 \, \text{K} $

Final Temperature of Sink

When the sink's temperature is lowered by 65 K:

$ T_2 - 65 = 325 - 65 = 260 \, \text{K} $

The initial temperature of the sink is 325 K, and the final temperature of the sink is 260 K.

To determine the minimum power output of the motor needed to prevent ice melting in a refrigerator, we must use the concept of the Coefficient of Performance (COP) for a refrigeration system.

The COP is defined as:

$ \omega = \frac{T_2}{T_1 - T_2} $

Where:

$ T_2 = 273 $ K (temperature inside the refrigerator, equivalent to 0°C)

$ T_1 = 293 $ K (external temperature, equivalent to 20°C)

Plugging in the values:

$ \omega = \frac{273}{293 - 273} = \frac{273}{20} $

Next, calculate the rate of ice melting. Given that ice melts at the rate of 2 kg per hour, convert this to grams per second:

$ \text{Rate of mass of ice melting per second} = \frac{2 \times 1000}{3600} = \frac{5}{9} \text{ g/s} $

To prevent the ice from melting, the heat that needs to be expelled per second, $ Q_2 $, is determined using the latent heat of fusion for ice, $ 80 \, \text{cal/g} $:

$ Q_2 = \left(\frac{5}{9}\right) \times 80 = \frac{400}{9} = 44.44 \, \text{cal/s} $

Convert $ Q_2 $ from calories to Joules (since $ 1 \, \text{cal} = 4.186 \, \text{J} $):

$ Q_2 = 44.44 \times 4.186 = 186 \, \text{J/s} $

Given this heat must be extracted to prevent melting, and using the COP:

$ \omega = \frac{Q_2}{W} \implies W = \frac{Q_2}{\omega} = \frac{186}{\left(\frac{273}{20}\right)} = 13.6 \, \text{J} $

Thus, in an ideal refrigerator, the minimum power output of the motor required is:

$ P_{\text{min}} = \frac{W}{1 \, \text{s}} = 13.6 \, \text{J/s} = 13.6 \, \text{W} $

Therefore, the minimum necessary power output is 13.6 W.

The efficiency of a Carnot engine is expressed as:

$ \eta = 1 - \frac{T_2}{T_1} $

where:

$ T_1 $ is the temperature of the source,

$ T_2 $ is the temperature of the sink, and

$ T_1 > T_2 $.

Case 1:

For a Carnot engine operating between 800 K and 500 K, the efficiency is:

$ \eta_1 = 1 - \frac{500}{800} $

Case 2:

For another Carnot engine with unknown source temperature $ x $ and a sink at 600 K, the efficiency is:

$ \eta_2 = 1 - \frac{600}{x} $

Given that the efficiencies are the same in both cases:

$ \begin{aligned} &\eta_1 = \eta_2 \\ &1 - \frac{600}{x} = 1 - \frac{500}{800} \\ &\frac{600}{x} = \frac{5}{8} \end{aligned} $

Solving for $ x $:

$ \begin{aligned} x &= \frac{600 \times 8}{5} \\ x &= \frac{4800}{5} \\ x &= 960 \, \text{K} \end{aligned} $

The rms velocity (root mean square velocity) of gas molecules is given by the formula:

$ v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}} $

Where:

$ R $ is the Universal gas constant

$ T $ is the temperature of the gas in Kelvin

$ M $ is the molar mass of the gas

For a constant $ R $ and $ M $, the rms velocity is proportional to the square root of the temperature:

$ v_{\mathrm{rms}} \propto \sqrt{T} $

Given a change in temperature from $ T' = 27^{\circ} \mathrm{C} $ to $ T'' = 90^{\circ} \mathrm{C} $, we first convert these temperatures to Kelvin:

$ T' = 273 + 27 = 300 \, \text{K} $

$ T'' = 273 + 90 = 363 \, \text{K} $

The ratio of the rms velocities at these temperatures is:

$ \frac{v_{\text{rms,new}}}{v_{\text{rms,old}}} = \sqrt{\frac{T''}{T'}} = \sqrt{\frac{363}{300}} $

This calculation gives:

$ \frac{v_{\text{rms,new}}}{v_{\text{rms,old}}} = \sqrt{1.21} \approx 1.1 $

The percentage increase in the rms velocity of the gas molecules is calculated by:

$ \left( \frac{v_{\text{rms,new}} - v_{\text{rms,old}}}{v_{\text{rms,old}}} \right) \times 100 = (1.1 - 1) \times 100 = 0.1 \times 100 = 10\% $

Thus, the increase in the rms velocity of the gas molecules is 10%.

According to Stefan's law, the rate of cooling is proportional to the difference between the fourth powers of the object's temperature and the ambient temperature:

$ H \propto \left(T^4 - T_0^4\right) $

Given:

Ambient temperature, $ T_0 = 300 \, \text{K} $

Rate of cooling at $ 600 \, \text{K} $ is $ H $

We are asked to find the rate of cooling at $ 900 \, \text{K} $.

Using Stefan's law, compare the rates of cooling:

$ \frac{H'}{H} = \frac{(900)^4 - (300)^4}{(600)^4 - (300)^4} $

Calculate:

$ 900^4 - 300^4 $

$ 600^4 - 300^4 $

$ \Rightarrow \frac{H'}{H} = \frac{9^4 - 3^4}{6^4 - 3^4} $

Simplify the expression:

$ 9^4 = 6561 $

$ 3^4 = 81 $

$ 6^4 = 1296 $

Substitute these values back:

$ \frac{H'}{H} = \frac{6561 - 81}{1296 - 81} = \frac{6480}{1215} = \frac{16}{3} $

Thus, the rate of cooling at $ 900 \, \text{K} $ is:

$ H' = \frac{16}{3} H $

Explanation:

For a Carnot cycle, the efficiency ($\eta$) of the heat engine is defined as:

$ \eta = 1 - \frac{T_2}{T_1} $

where $ T_1 $ is the absolute temperature of the heat source (in Kelvin), and $ T_2 $ is the absolute temperature of the heat sink (in Kelvin).

Convert the given temperatures from Celsius to Kelvin:

$ T_1 = 127^\circ \mathrm{C} + 273 = 400 \, \mathrm{K} $

$ T_2 = 27^\circ \mathrm{C} + 273 = 300 \, \mathrm{K} $

Plug these values into the efficiency formula:

$ \eta = 1 - \frac{300}{400} = 1 - 0.75 = 0.25 $

This means the efficiency is $ \frac{1}{4} $.

The efficiency also relates to the work output ($ W_{\text{out}} $) and the heat absorbed ($ W_{\text{absorb}} $) as follows:

$ \eta = \frac{W_{\text{out}}}{W_{\text{absorb}}} $

Given that the heat absorbed ($ W_{\text{absorb}} $) is $ 5 \times 10^4 $ cal, we can calculate the work output:

$ W_{\text{out}} = \frac{1}{4} \times 5 \times 10^4 = 1.25 \times 10^4 \, \mathrm{cal} $

Thus, the amount of heat converted to work is $ 1.25 \times 10^4 $ cal.

To find the value of $\gamma$ for the mixture of two gases, we use the formula:

$ \frac{n_1 + n_2}{\gamma_{\text{mix}} - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1} $

For our specific case:

$n_1 = 1$, $\gamma_1 = \frac{7}{5}$

$n_2 = 1$, $\gamma_2 = \frac{4}{3}$

Substituting these into the formula gives:

$ \frac{1 + 1}{\gamma_{\text{mix}} - 1} = \frac{1}{\frac{7}{5} - 1} + \frac{1}{\frac{4}{3} - 1} $

This simplifies to:

$ \frac{2}{\gamma_{\text{mix}} - 1} = \frac{1}{\frac{2}{5}} + \frac{1}{\frac{1}{3}} $

Calculating further:

$ \frac{2}{\gamma_{\text{mix}} - 1} = \frac{5}{2} + 3 = \frac{5}{2} + \frac{9}{3} = \frac{5}{2} + \frac{6}{2} = \frac{11}{2} $

Now, solving for $\gamma_{\text{mix}}$:

$ \gamma_{\text{mix}} - 1 = \frac{2}{\frac{11}{2}} = \frac{4}{11} $

So, $\gamma_{\text{mix}}$ is:

$ \gamma_{\text{mix}} = 1 + \frac{4}{11} = \frac{15}{11} $

A Carnot heat engine with an efficiency of 10% can be transformed into a refrigerator. To find the coefficient of performance (COP) of this refrigerator, we use the relationships between efficiency and temperature in a Carnot cycle.

The efficiency $\eta$ of a Carnot engine is given by:

$ \eta = 1 - \frac{T_2}{T_1} $

Rearranging this formula, we find:

$ \frac{T_1}{T_2} = \frac{1}{1 - \eta} \tag{i} $

For a refrigerator, the coefficient of performance $\omega$ is defined as:

$ \omega = \frac{T_2}{T_1 - T_2} $

Substituting equation (i) into this, we have:

$ \omega = \frac{1}{\left(\frac{T_1}{T_2}\right) - 1} = \frac{1}{\left[\frac{1}{1 - \eta}\right] - 1} = \frac{1 - \eta}{\eta} $

Plugging in the given efficiency ($\eta = 0.1$):

$ \omega = \frac{1 - 0.1}{0.1} = \frac{0.9}{0.1} = 9 $

Therefore, the coefficient of performance of the refrigerator is 9.

To determine the relationship between the root-mean-square (rms) velocity of gas molecules and their mass, we use the formula:

$ v_{\mathrm{rms}} = \sqrt{\frac{3 R T}{m}} $

In this equation:

$ v_{\mathrm{rms}} $ is the rms velocity,

$ R $ is the gas constant,

$ T $ is the temperature,

$ m $ is the mass of the gas molecules.

From the equation, it's clear that:

$ v_{\mathrm{rms}} \propto \frac{1}{\sqrt{m}} $

This means the rms velocity of gas molecules is inversely proportional to the square root of their mass. As the mass $ m $ of the molecules increases, the rms velocity decreases, and vice versa.

The dimensions of various thermal properties are given and described as follows:

Thermal Conductivity (K):

Formula: $ K = \frac{Q \cdot d}{A \cdot \Delta T} $

Dimensional Analysis:

$ \left[ \frac{(ML^2T^{-3}) (L)}{(L^2)(K)} \right] = [MLT^{-3}K^{-1}] $

Boltzmann Constant (k):

Defined as the ratio of energy to temperature:

Dimensional Analysis:

$ \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}] $

Latent Heat (L):

Defined as the ratio of energy to mass:

Dimensional Analysis:

$ \frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}] $

Specific Heat (C):

Defined as the ratio of energy to the product of mass and temperature:

Dimensional Analysis:

$ \frac{[ML^2T^{-2}]}{[M][K]} = [M^0L^2T^{-2}K^{-1}] $

These analyses help in determining the correct dimension pairings for each property.

Given:

Length of the rod, $ x = 20 \, \text{cm} = 0.2 \, \text{m} $

Temperatures at the ends:

$ T_1 = 0^{\circ} \text{C} = 273 \, \text{K} $

$ T_2 = 100^{\circ} \text{C} = 373 \, \text{K} $

Cross-sectional area, $ A = 4 \times 10^{-4} \, \text{m}^2 $

Mass of ice melted, $ m = 5 \, \text{g} $

Time, $ t = 1 \, \text{min} = 60 \, \text{s} $

Latent heat of fusion, $ L_{\text{ice}} = 80 \, \text{cal/g} = 336 \, \text{J/g} $

Calculation:

Total Heat Required to Melt the Ice:

$ Q = m \times L_{\text{ice}} = 5 \times 336 = 1680 \, \text{J} $

Rate of Heat Transfer:

$ \frac{Q}{t} = \frac{1680}{60} = 28 \, \text{J/s} $

Thermal Conductivity ($ K $):

$ K = \frac{\Delta Q \cdot x}{A \cdot \Delta T} = \frac{28 \times 0.2}{4 \times 10^{-4} (373 - 273)} $

$ K = \frac{1.4 \times 10^4}{100} $

$ K = 1.4 \times 10^2 $

$ K = 140 \, \text{W/mK} $

Given:

Initial work done by the gas, $ W_1 = W $

Final volume for the first process, $ V_{f_1} = 2V $

Initial volume for the first process, $ V_{i_1} = V $

Temperature for the first process, $ T_1 = T $

Number of moles for the first process, $ n = 2 $

The work done by an ideal gas during an isothermal expansion is given by the equation:

$ W = n \cdot R \cdot T \ln \left(\frac{V_f}{V_i}\right) $

Substituting the given values for the first process:

$ W_1 = 2 \cdot R \cdot T \ln \left(\frac{2V}{V}\right) = 2 \cdot R \cdot T \ln 2 $

For the second process:

Final volume, $ V_{f_2} = 8V $

Initial volume, $ V_{i_2} = V $

Temperature, $ T_2 = \frac{T}{2} $

Number of moles, $ n_2 = 4 $

The work done for the second process is:

$ W_2 = n_2 \cdot R \cdot T_2 \ln \left(\frac{V_{f_2}}{V_{i_2}}\right) $

Substituting the values:

$ W_2 = 4 \cdot R \cdot \frac{T}{2} \ln \left(\frac{8V}{V}\right) $

$ W_2 = 2 \cdot R \cdot T \ln \left(2^3\right) $

$ W_2 = 3 \cdot 2 \cdot R \cdot T \ln 2 $

This simplifies to:

$ W_2 = 3 \cdot W $

Hence, the work done by the ideal gas in the second process is $ 3W $.

To analyze the increase in internal energy when heat is absorbed by a monoatomic gas:

First, we know that the heat absorbed, $ Q $, is given as 40 J.

For a monoatomic ideal gas, the specific heat capacities are:

$ C_p = \frac{5R}{2} $

$ C_V = \frac{3R}{2} $

The relationship for heat transfer in terms of $ C_p $ is:

$ Q = n C_p \Delta T $

Substituting in known values:

$ 40 = n \cdot \frac{5R}{2} \cdot \Delta T $

From this, we obtain:

$ nR \Delta T = 16 \quad (i) $

Next, we determine the increase in internal energy, $ \Delta U $, which is given by:

$ \Delta U = n C_V \Delta T $

$ \Delta U = n \cdot \frac{3}{2} R \Delta T $

$ \Delta U = \frac{3}{2} \cdot n R \Delta T $

Using Equation (i), we substitute for $ nR \Delta T $:

$ \Delta U = \frac{3}{2} \times 16 $

This simplifies to:

$ \Delta U = 24 \, \text{J} $

Thus, the increase in the internal energy of the gas is 24 J.

The efficiency of a Carnot engine is calculated using the formula:

$ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} $

Given that the absolute temperature of the source ($T_{\text{source}}$) is 25% more than that of the sink ($T_{\text{sink}}$), we can express this relationship as:

$ T_{\text{source}} = T_{\text{sink}} + \frac{25 \times T_{\text{sink}}}{100} $

Simplifying, this becomes:

$ T_{\text{source}} = 1.25 \times T_{\text{sink}} $

Substituting into the efficiency equation, we have:

$ \eta = \left(1 - \frac{T_{\text{sink}}}{1.25 \times T_{\text{sink}}}\right) \times 100\% $

This simplifies to:

$ \eta = \left(1 - \frac{1}{1.25}\right) \times 100\% $

$ \eta = (1 - 0.8) \times 100\% $

$ \eta = 0.2 \times 100\% $

Thus, the efficiency of the Carnot engine is:

$ \eta = 20\% $

Given:

Molar heat capacity at constant pressure for a diatomic gas, $ C_p = C $

We need to find the molar heat capacity at constant volume for a monoatomic gas, $ C_V $.

Calculations:

For a Diatomic Gas:

For a diatomic gas, the molar heat capacity at constant volume is:

$ C_V = \frac{5}{2} R $

Therefore, the relation between $ C_p $ and $ C_V $ is:

$ C_p = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R $

Since $ C_p = C $, we have:

$ \frac{7}{2}R = C $

Solving for $ R $, we get:

$ R = \frac{2C}{7} \tag{i} $

For a Monoatomic Gas:

The molar heat capacity at constant volume for a monoatomic gas is:

$ C_V = \frac{3}{2} R $

Substituting the value of $ R $ from Equation (i):

$ C_V = \frac{3}{2} \cdot \frac{2C}{7} = \frac{3C}{7} $

Therefore, the molar specific heat capacity of a monoatomic gas at constant volume is $ \frac{3C}{7} $.

Given, $s_w=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$

$\begin{aligned} & s_{\text {lece }}=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1} \\ & L_f=80 \mathrm{cal} \mathrm{g}^{-1} \end{aligned}$

Mass of ice, $m_{\text {cee }}=5 \mathrm{~g}$

Mass of water, $m_w=20 \mathrm{~g}$

Let final temperature of the mixture be $T^{\circ} \mathrm{C}$.

According to principle of calorimetry,

Heat lost by water $=$ Heat gained by ice

$\begin{aligned} & m_w s_w(35-T)=m_{\text {cce }} s_{\text {ce }}[0-(-30)] \\ & +\mathrm{m}_{i c c} \times L_f+m_{\text {kec }} \times S_w \times T \\ & \Rightarrow 20 \times 1 \times(35-T)=5 \times 0.5 \times 30+5 \times 80+5 \times 0.5 \times T \\ \end{aligned}$

On solving, we get

$T=9^{\circ} \mathrm{C}$

Given :

• Diameter of the sphere, $D$

Initial surface area, $A$ is:

$ A = 4 \pi R^2 $

where $R = \frac{D}{2}$.

So,

$ A = 4 \pi \left(\frac{D}{2}\right)^2 = \pi D^2 \quad \text{(i)} $

After heating by temperature $\delta T$, the new surface area, $A'$, is:

$ A' = 4 \pi \left(\frac{D'}{2}\right)^2 = \pi \left(D'\right)^2 \quad \text{(ii)} $

Here, $D'$ is the new diameter.

Using the equation for linear expansion:

$ D' = D(1 + \alpha \delta T) \quad \text{(iii)} $

Substituting the value of $D'$ from equation (iii) into equation (ii), we get:

$ \begin{aligned} A' & = \pi D^2 (1 + \alpha \delta T)^2 \\ & = \pi D^2 \left(1 + \alpha^2 \delta T^2 + 2 \alpha \delta T\right) \quad \text{(iv)} \end{aligned} $

Simplifying further:

$ \begin{aligned} A' & = \pi D^2 \left(1 + \alpha^2 \delta T^2 + 2 \alpha \delta T\right) - \pi D^2 \\ & = \pi D^2 \left[1 + \alpha^2 \delta T^2 + 2 \alpha \delta T - 1\right] \\ & = \pi D^2 \left[\alpha^2 \delta T^2 + 2 \alpha \delta T\right] \end{aligned} $

The change in surface area is:

$ \begin{aligned} \Delta A & = A' - A \\ & = \pi D'^{ 2} - \pi D^{ 2 } \\ & = \pi D^2 \left(\alpha \delta T (\alpha \delta T + 2)\right) \end{aligned} $

To find the energy exhausted by the Carnot engine, we need to understand the relationship between the work done, the heat absorbed, and the heat exhausted in a Carnot cycle.

The efficiency ($\eta$) of a Carnot engine operating between two temperatures $T_{h}$ (hot reservoir) and $T_{c}$ (cold reservoir) is given by the formula:

$\eta = 1 - \frac{T_{c}}{T_{h}}$

In this case, the temperatures are:

$T_{h} = 400 \, \text{K}$

$T_{c} = 300 \, \text{K}$

Substituting these values into the efficiency formula, we get:

$\eta = 1 - \frac{300}{400}$

$\eta = 1 - 0.75$

$\eta = 0.25$

The efficiency of the Carnot engine is 0.25 (or 25%). This means that 25% of the heat absorbed (Q_h) is converted into work (W), and the rest is exhausted as waste heat (Q_c).

We know that the work done by the engine (W) is 400 J. Using the efficiency, we can find the heat absorbed (Q_h):

$W = \eta \cdot Q_{h}$

$400 = 0.25 \cdot Q_{h}$

$Q_{h} = \frac{400}{0.25}$

$Q_{h} = 1600 \, \text{J}$

The heat absorbed by the engine is 1600 J. To find the energy exhausted (Q_c), we use the relationship:

$Q_{h} = W + Q_{c}$

Substituting the known values, we get:

$1600 = 400 + Q_{c}$

$Q_{c} = 1600 - 400$

$Q_{c} = 1200 \, \text{J}$

Therefore, the energy exhausted by the engine is 1200 J.

The correct answer is:

Option B: 1200 J

To solve this problem, we need to understand the relationship between the slopes of isothermal and adiabatic processes in a pressure-volume ($p-V$) diagram for a gas, as well as the heat capacity ratio (also known as the adiabatic index $\gamma$).

For an isothermal process (constant temperature), the pressure and volume relationship is given by Boyle's law:

$ pV = \text{constant} $

Differentiating the above with respect to volume, we get:

$ \frac{dp}{dV} = -\frac{p}{V} $

Thus, the slope of the isothermal $p-V$ graph, $S_I$, is:

$ S_I = -\frac{p}{V} $

For an adiabatic process (no heat exchange), the relationship between pressure and volume is given by:

$ pV^\gamma = \text{constant} $

where $\gamma$ is the heat capacity ratio. Differentiating the above with respect to volume, we get:

$ \frac{dp}{dV} = -\gamma \frac{p}{V} $

Thus, the slope of the adiabatic $p-V$ graph, $S_A$, is:

$ S_A = -\gamma \frac{p}{V} $

From the above, we can see that the ratio of $S_I$ to $S_A$ is:

$ \frac{S_I}{S_A} = \frac{-\frac{p}{V}}{-\gamma \frac{p}{V}} = \frac{1}{\gamma} $

Given that the heat capacity ratio $\gamma = \frac{3}{2}$, we substitute this value into the ratio:

$ \frac{S_I}{S_A} = \frac{1}{\frac{3}{2}} = \frac{2}{3} $

Thus, the correct answer is:

Option B: $\frac{2}{3}$

The number of rotational degrees of freedom of a diatomic molecule is an important concept in molecular physics, which helps in understanding how the molecule can rotate in space.

A diatomic molecule consists of two atoms connected by a bond. Due to this structure, the molecule can rotate around two axes perpendicular to the bond axis. However, rotation around the bond axis itself does not count as a rotational degree of freedom for rigid diatomic molecules because the moment of inertia around this axis is negligible.

Therefore, the diatomic molecule has:

• Two rotational degrees of freedom: rotation around two axes perpendicular to the bond axis.

Based on this explanation, the correct answer is:

Option C: 2

Given:

• Coefficient of linear expansion, $\alpha = 1 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1}$


• Initial temperature, $T_1 = 25^{\circ} \mathrm{C}$


• Final temperature, $T_2 = -15^{\circ} \mathrm{C}$

Firstly, calculate the change in temperature:

$ \Delta T = T_1 - T_2 = 25 - (-15) = 40^{\circ} \mathrm{C} $

Next, determine the change in length ($\Delta L$) with respect to the original length (L):

$ \Delta L = L \alpha \Delta T $

Thus, the fractional change in length is:

$ \frac{\Delta L}{L} = \alpha \Delta T $

$ \frac{\Delta L}{L} = 1 \times 10^{-5} \times 40 $

$ \frac{\Delta L}{L} = 4 \times 10^{-4} $

To express this as a percentage change in the measurement of length:

$ \frac{\Delta L}{L} \times 100 = 4 \times 10^{-4} \times 100 $

$ \frac{\Delta L}{L} \times 100 = 0.04\% $

Therefore, the percentage error in the measurement of length is 0.04%.

Hint : Work done by adiabatic expansion $=-$ (Change in internal energy) $=-(-30 \mathrm{~J})=30 \mathrm{~J}$