Heat and Thermodynamics

Initial temperature of the sink, $T_2=250 \mathrm{~K}$

Initial efficiency of carnot engine,

$\eta_1=25 \%=0.25$

If $T_1$ be the temperature of the sink, then

$\begin{aligned} \eta_1 & =1-\frac{T_2}{T_1} \\ 0.25 & =1-\frac{250}{T_1} \Rightarrow \frac{250}{T_1}=0.75 \\ \Rightarrow \quad T_1 & =\frac{1000}{3} \mathrm{~K} \end{aligned}$

When efficiency is increased by $50 \%$

i.e $\eta_2=50 \%=0.5$

Then if $T_2^{\prime}$ be temperature of sink, then

$\begin{array}{rlrl} & \eta_2 & =1-\frac{T_2^{\prime}}{T_1} \\ \Rightarrow & & 0.5 & =1-\frac{T_2^{\prime}}{1000 / 3} \\ \Rightarrow & & \frac{T_2^{\prime}}{1000 / 3} & =1-0.5=0.5=\frac{1}{2} \\ \Rightarrow & & T_2^{\prime} & =\frac{1}{2} \times \frac{1000}{3}=\frac{500}{3} \mathrm{~K} \end{array}$

Here we see that to increase the efficiency of carnot engine temperature of the sink should be decreased, but in question, asked about increasing temperature of sink to increase efficiency, which is not possible. Hence, no option is correct.

In non-rigid diatomic molecule with an additional vibrational mode, each molecule has additional energy equal to $2 \times \frac{1}{2} k_B T=k_B T$, because a vibrational frequency has both kinetic and potential energy modes.

$\begin{aligned} \therefore \quad U & =\left(\frac{5}{2} k_B T+k_B T\right) N_A=\frac{7}{2} k_B N_A T \\ & =\frac{7}{2} R T \quad\left[R=k_B N_A\right] \end{aligned}$

$\begin{aligned} \therefore \quad C_v & =\frac{d U}{d T}=\frac{d}{d T}\left(\frac{7}{2} R T\right)=\frac{7}{2} R \\ C_p & =C_V+R=\frac{7}{2} R+R=\frac{9}{2} R \\ \frac{C_p}{C_V} & =\frac{9 / 2 R}{7 / 2 R} \Rightarrow \frac{C_p}{C_V}=\frac{9}{7} \\ \Rightarrow \quad 49 C_p^2 & =81 C_V^2 \end{aligned}$

Given surface area of sphere, $A=4 \mathrm{~m}^2$

Temperature, $T=400 \mathrm{~K}$

Emissivity, $\varepsilon=0.5$

Environment temperature,

$T_0=200 \mathrm{~K}$

Stefan Boltzmann constant, $\sigma=5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^4$

According to Stefan Boltzmann law, rate of energy exchange of the sphere

$\begin{aligned} E & =\varepsilon \sigma A\left(T^4-T_0^4\right) \\ & =0.5 \times 5.67 \times 10^{-8} \times 4\left(400^4-200^4\right) \\ & =27216 \times 10^3 \mathrm{~W}=2721.6 \mathrm{~W} \end{aligned}$

Efficiency of carnot engine,

$\eta=1-\frac{T_2}{T_1}\quad \text{... (i)}$

Where, $T_2=$ Temperature of sink and $T_1=$ temperature of source According to first condition,

$\begin{aligned} & T_2=27+273=300 \mathrm{~K} \\ & \eta=40 \%=0.4 \end{aligned}$

From Eq. (i), $0.4=1-\frac{300}{T_1}$

$\Rightarrow \quad \frac{300}{T_1}=0.6 \Rightarrow T_1=\frac{300}{0.6}=500 \mathrm{~K}$

$\Rightarrow \quad T_1=500 \mathrm{~K}$

Again, when, $\eta=50 \%=0.5$, then

$\begin{aligned} & \qquad 0.5=1-\frac{300}{T_1^{\prime}} \\ & \Rightarrow \quad \frac{300}{T_1^{\prime}}=0.5 \Rightarrow T_1^{\prime}=600 \mathrm{~K} \\ & \begin{aligned} & \therefore \text { Increase of source temperature }=T^{\prime}-T_1 \\ &=600-500=100 \mathrm{~K} \end{aligned} \end{aligned}$

given, power of engine, $P=20 \mathrm{~kW}$

$=2 \times 10^4 \mathrm{~W}$

Efficiency, $\eta=40 \%=0.4$

Time, $t=1$ hour $=60 \times 60 \mathrm{~s}$

Energy of car, $E=$ Power $\times$ Time

$=2 \times 10^4 \times 60 \times 60=7.2 \times 10^7 \mathrm{~J}$

$\begin{aligned} & \text { Energy generated by fuel Combustion }=\frac{E}{\eta} \\ & =\frac{7.2 \times 10^7}{0.4}=18 \times 10^7 \mathrm{~J} \\ & =18 \times 10^4 \times 10^3 \mathrm{~J}=18 \times 10^4 \mathrm{~kJ}=180000 \mathrm{~kJ} \end{aligned}$

A diatomic molecule has one normal mode of vibration, since it can only stretch or compress the single bond. Hence, it has only one vibrational degree of freedom.

Given,

Volume of glass vessel and liquid $=V_o$

Change in temperature $=\Delta T$

Linear expansion of glass $=\alpha_g$

Volume expansion of liquid $=\gamma_l$

Let final volume of glass and liquid be $V_g$ and $V_l$, then

$\begin{aligned} & V_g=V_o\left(1+3 \alpha_g \Delta T\right) \\ & V_l=V_o\left(1+\gamma_l \Delta T\right) \end{aligned}$

Volume of liquid flowing out,

$V_l-V_g=V_0\left(1+\gamma_l \Delta T\right)-V_0\left(1+3 \alpha_g \Delta T\right)=V_o \Delta T\left(\gamma_l-3 \alpha_g\right)$

Given, sink temperature, $T_2=27^{\circ} \mathrm{C}=300 \mathrm{~K}$

Case I Efficiency, $\eta_1=40 \%$

Case II Efficiency, $\eta_2=60 \%$

Let temperature of source in two cases be $T_1$ and $T_1^{\prime}$.

As we know that,

$\begin{aligned} \eta_1 & =1-\frac{T_2}{T_1} \\ \Rightarrow \quad \frac{40}{100} & =1-\frac{300}{T_1} \end{aligned}$

$\begin{aligned} & \Rightarrow \quad T_1=500 \mathrm{~K} \\ & \text { and } \quad \eta_2=1-\frac{T_2}{T_1^{\prime}} \\ & \Rightarrow \quad \frac{60}{100}=1-\frac{300}{T_1^{\prime}} \Rightarrow \frac{300}{T_1^{\prime}}=\frac{40}{100} \\ & \Rightarrow \quad T_1^{\prime}=750 \mathrm{~K} \\ & \therefore \text { Change is temperature, } \Delta T=T_1^{\prime}-T_1 \\ & =750-500=250 \mathrm{~K} \\ & \end{aligned}$

Given, nature of gas = Diatomic (at constant pressure)

As we know that,

Internal energy, $U=n C_V \Delta T$ ...... (i)

Heat energy, $Q=n C_p \Delta T$ ...... (ii)

where, $n=$ number of mole

$C_V=$ specific heat at constant volume

$C_p=$ specific heat at constant pressure

and $C_p=C_V+R$

If $f=$ degree of freedom, then

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

Now, dividing Eq. (i) by Eq. (ii), we get

$\Rightarrow \quad \frac{U}{Q}=\frac{n C_V \Delta T}{n C_p \Delta T}=\frac{C_V}{C_p}=\frac{5 R / 2}{7 R / 2}=\frac{5}{7}$

According to given $p$ - $V$ diagram,

As we know that work is equal to area under $p$-$V$ graph.

$\therefore \quad W_A > W_B > W_C > W_D$

and by using 1st law of thermodynamics,

$\Delta Q=\Delta U+\Delta W$

Here, internal energy, $\Delta U=0$

$\because$ Initial and final positions are same.

$\therefore$ Heat $(\Delta Q)$ sequence will be

$\Delta Q_A > \Delta Q_B > \Delta Q_C > \Delta Q_D $

Hence, $Q_A-Q_D=W_A-W_D$

The gas has its temperature increased from 27$^\circ$C to 127$^\circ$C.

First, change the temperatures from degrees Celsius to Kelvin:

$T_1 = 27 + 273 = 300\,K$

$T_2 = 127 + 273 = 400\,K$

The root mean square (rms) speed of a gas, $v_{rms}$, is related to temperature by the formula:

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

Here, $R$ is the gas constant and $M$ is the molar mass. Because $R$ and $M$ stay the same, $v_{rms}$ depends only on temperature. We can say:

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

So, the ratio of the rms speeds at $T_1$ and $T_2$ is:

$ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \frac{2}{\sqrt{3}} $

To find the percentage increase in $v_{rms}$, use:

$ \text{Percentage increase} = \left(\frac{v_2 - v_1}{v_1}\right) \times 100 $

Substitute the values:

$ \left(\frac{v_2}{v_1} - 1\right) \times 100 = \left(\frac{2}{\sqrt{3}} - 1\right) \times 100 $

This gives about 15.5% increase in $v_{rms}$.

Specific heat is the amount of heat required to raise the temperature of water by $1^{\circ} \mathrm{C}$.

For boiling of water to turn into steam, $\Delta T=0$

Since, $\Delta Q=n C_p \Delta T$

If $\Delta T=0, C_p=\frac{\Delta Q}{\Delta T}=\frac{\Delta Q}{0}=\infty$

Given,

Volume change, $\frac{\Delta V}{V}=\frac{0.12}{100}=12 \times 10^{-4}$

Change in temperature, $\Delta T=20 \Upsilon C$

Since, $\Delta V / V=3 \alpha \Delta T$

where, $\alpha$ is coefficient of linear expansion.

$\therefore \quad \alpha=\frac{\Delta V}{V} \times \frac{1}{3 \Delta T} \Rightarrow \alpha=\frac{12 \times 10^{-4}}{3 \times 20}=2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$

As we know that,

In case of isothermal condition, T = constant

By ideal gas law,

$pV = nRT$

$\Rightarrow \quad pV=$ constant

$\Rightarrow \quad p\propto \frac{1}{V}$

$\therefore p$-$V$ graph is given as:

According to given graph,

As we know that,

$C=\frac{R}{\gamma-1}-\frac{R}{m-1}$ ..... (i)

where, $C_V$ is heat coefficient at constant volume, R is universal gas constant = 8.314 J K$^{-1}$, $\gamma$ is adiabatic constant and m is polytropic constant.

For path $A B, p V^2=$ constant

$\therefore \quad m=2$

and $\gamma_{\text {monoatomic }}=5 / 2$

Put these values in Eq (i), we get

$C=\frac{R}{\frac{5}{3}-1}-\frac{R}{2-1}=\frac{3 R}{2}-R=\frac{R}{2}$

For path $B C$, density is fixed

$\therefore$ Volume will remain unchanged.

Since, $U=n C_V \Delta T$

where, $U$ is internal energy and $C_V$ (monoatomic gas) $=\frac{3}{2} R$.

Isochoric process, $\Delta V=0$

But temperature is decreasing and $C_V=\frac{3}{2} R$

$\therefore$ Option (c) is incorrect.

For path $C A$ (isothermal)

$W=n R T \ln \frac{p_2}{p_1}=n R T \ln \frac{4 p_0}{p_0}=n R T \ln 4$

and $\quad U_0=\frac{3}{2} n R T$

$\therefore \quad W=\frac{2}{3} U_0(\ln 4)$

As we know that,

Kinetic energy, per unit volume, $E=\frac{1}{2} \rho V^2$ ..... (i)

and pressure per unit volume, $p=\frac{1}{3} \rho V^2$ ..... (ii)

From Eqs. (i) and (ii), we get

$\therefore \quad p=\frac{1}{3} 2 E \Rightarrow p=\frac{2}{3} E$

Hence, $p \propto E$.

As we know that, during expansion due to heating, change in volume $\Delta V=V_0 \gamma \Delta T$

Here, $V_0, \gamma$ and $\Delta T$ are initial volume, volumetric expansion coefficient and change in temperature.

Since, density $(\rho)=\frac{\operatorname{mass}(m)}{\text { volume }(V)}$

As volume increases by $\Delta V$.

$\therefore$ Density decreases.

As we know that,

(A) Since, time period $(T)=2 \pi \sqrt{l / g}$ where, $l$ is length and $g$ is gravity.

$\therefore \quad \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta l}{l}$ ..... (i)

and by using concept of linear expansion,

$\frac{\Delta l}{l}=\alpha \Delta T$ ..... (ii)

Substituting in Eq. (i), we get

$\frac{\Delta T}{T}=\frac{1}{2} \alpha \Delta T$

$(\mathrm{A} \rightarrow 4)$

(B) Now, from Eq. (ii)

$\begin{aligned} \Delta l & =l \alpha \Delta T \\ \Rightarrow \quad l_0-l & =l \alpha \Delta T \Rightarrow l_0=l+l \alpha \Delta T \\ \Rightarrow \quad l_0 / l & =l+\alpha \Delta T \end{aligned}$

(B) $\rightarrow(3)$

$\begin{array}{ll} \text {(C)} \because p \Delta V=n R T \quad .....\text{(i)}\\ \text { and } \Delta V=V_0 \gamma \Delta T \\ \Rightarrow \Delta V \propto \gamma \\ \Rightarrow \gamma \propto T\quad \text {[Using Eq. (i)]}\end{array}$

(C) $\rightarrow(2)$

(D) Since, $Y=\frac{F l}{A \Delta l} \Rightarrow \frac{\Delta l}{l}=\frac{F}{A Y}=\alpha \Delta T$

(D) $\rightarrow$ (1)

$\therefore \quad \mathrm{A} \rightarrow 4, \mathrm{~B} \rightarrow 3, \mathrm{C} \rightarrow 2, \mathrm{D} \rightarrow \mathrm{l}$ is correct.

As we know that,

Irreversible process is the phenomenon which can’t be reversed back to initial condition.

Since, conduction is the phenomenon in which heat flows from higher temperature to lower temperature.

$\therefore$ Conduction is irreversible process.

Given, internal energy = U

Temperature = T

Since, $U=\frac{3}{2}KT$

$\therefore \quad U \propto T$

Given, coefficient of performance $(\mathrm{COP})=0.25$

Heat coming out, $Q_c=250 \mathrm{~J}$

Since,

$\begin{aligned} & \operatorname{COP}=\frac{Q_L}{\operatorname{Work}(W)} \\ & \Rightarrow \quad \mathrm{COP}=\frac{Q_L}{Q_H-Q_L} \Rightarrow \frac{Q_H-Q_L}{Q_L}=\frac{100}{25}=4 \\ & \Rightarrow \quad \frac{Q_H}{Q_L}-1=4 \Rightarrow \frac{Q_H}{Q_L}=5 \\ & \Rightarrow \quad Q_L=\frac{250}{5}=50 \\ & \therefore \quad W=Q_H-Q_L \\ & =250-50=200 \mathrm{~J} \\ \end{aligned}$

Given, mass of oxygen, $m_i=6 \mathrm{~g}$

Initial pressure $p_i=p$

Initial temperature $T_i=400 \mathrm{~K}$

Mass of oxygen leak $=m^{\prime}$

Final pressure, $p_f=\frac{p}{2}$

Final temperature, $T_f=300 \mathrm{~K}$

Since, $p V=\frac{m}{M} R T$

where, $m$ is given mass

$M$ is molar mass and $R$ is gas constant

Now, $\frac{p_i}{T_i m_i}=\frac{p_f}{T_f m_f}$

$\Rightarrow m_f=\left(\frac{p_f}{T_f}\right) \cdot\left(\frac{T_i m_i}{p_i}\right)=\left(\frac{p}{2 \times 300} \times \frac{400 \times 6}{p}\right)=4 \mathrm{~g}$

$\therefore$ Mass of oxygen leakes out $=6-4=2 \mathrm{~g}$

Given, initial and final length of rod is $l_1$ and $l_2$ i.e., $20 \mathrm{~cm}$ and $9 \mathrm{~cm}$ from $A$, and temperature at $A$ and $B$ is $T_A$ and $T_B$ i.e., $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$.

Now, since, heat flow rate $=\frac{d Q}{d t}=\frac{K A \Delta \theta}{\lambda}$

As the flow is steady,

$\begin{aligned} & \therefore & \frac{K A(100-\theta)}{9} & =\frac{K A(\theta-0)}{11} \\\\ & \Rightarrow & 1100-11 \theta & =9 \theta \\\\ &\Rightarrow & 1100 =20 \theta \\\\ & \therefore & \theta =55^{\circ} \mathrm{C} \end{aligned}$

Given, length of rods are $L$ and $2 L$.

Coefficient of linear expansion $\alpha$ and $2 \alpha$.

As, we know that,

$\Delta l=l_0 \alpha \Delta T$

where, $\Delta l$ is change in length,

$T$ is temperature

and $l_0$ is initial length

let final heat coefficient be $\alpha^{\prime}$

$\begin{array}{llrl} & \therefore & (L+2 L) \alpha^{\prime} \Delta T & =L \alpha \Delta T+2 L \cdot 2 \alpha \cdot \Delta T \\ \Rightarrow & & 3 \alpha^{\prime} & =\alpha+4 \alpha=5 \alpha \\ \Rightarrow & & \alpha^{\prime} & =5 \alpha / 3 \end{array}$

Given, heat absorbed and work done by two systems be $Q_1, Q_2$ and $W_1, W_2$ and as we know that, By using Ist law of thermodynamics,

$\Delta Q=\Delta U+\Delta W \Rightarrow \Delta U=\Delta Q-\Delta W$

where, $\Delta Q$ is heat change, $\Delta U$ is internal energy change i.e. state function and $\Delta W$ is work done.

$\begin{array}{lc} \therefore & \Delta Q_1-\Delta W_1=\Delta Q_2-\Delta W_2 \\ \text { or } & Q_1-W_1=Q_2-W_2 \end{array}$

Given, adiabatic constant $\gamma=1.5=3 / 2$

Let initial volume, $V_1=V$

$\therefore$ Final volume, $V_2=V / 4$

Initial and final pressure be $p_1$ and $p_2$.

$\begin{aligned} & \text { Since, } \quad p V^\gamma=\text { constant } \\ & \therefore \quad p_1 V_1^\gamma=p_2 V_2^\gamma \\ & \Rightarrow \quad p_1 / p_2=\left(V_2 / V_1\right)^\gamma=\left(\frac{V}{4 \times V}\right)^{3 / 2} \\ \end{aligned}$

$\Rightarrow p_1/p_2=\frac{1}{8}$

Hence, $p_2:p_1=8:1$

Given, initial temperature $T_1 =30^{\circ} \mathrm{C} =303 \mathrm{~K}$

Change in radius, $\Delta r=0.08 \mathrm{~mm}$

Internal diameter of cylinder $d_c=15 \mathrm{~cm}$

$\therefore$ Internal radius of cylinder, $r_c=7.5 \times 10^{-2} \mathrm{~m}$

Radius of piston, $r_p=(7.5-0.008) \times 10^{-2} \mathrm{~m}$

$=7.492 \times 10^{-2} \mathrm{~m}$

Linear expansion of cylinder $\alpha_C$ and piston $\alpha_p$ are $1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}$ and $1.6 \times 10^{-5} /{ }^{\circ} \mathrm{C}$.

Let, $\Delta r_C$ and $\Delta r_p$ is increase in radius of cylinder and piston respectively, for fully filled piston in cylinder.

By using equation of linear expansion,

$\begin{aligned} & r=r_0(1+\alpha \Delta T) \\ \therefore \quad & \Delta r_C=r_C \alpha_C \Delta T, \text { for cylinder } \end{aligned}$

New radius will be $r_C^{\prime}=r_C+\Delta r_C$

$\begin{aligned} & =r_C+r_C \alpha_C \Delta T=r_C\left(1+\alpha_C \Delta T\right) \\ & =7.5 \times 10^{-2}\left[1+1.2 \times 10^{-5}(T-303)\right] \end{aligned}$

Similarly, new radius of piston,

$r_p{ }^{\prime}=7.492 \times 10^{-2}\left[1+1.6 \times 10^{-5}(T-303)\right]$

Now, for fully fitted piston, $r_C=r_p$

$\begin{aligned} 7.5 \times 10^{-2}[1 & \left.+1.2 \times 10^{-5}(T-303)\right] \\ & =7.492 \times 10^{-2}\left[1+1.6 \times 10^{-5}(T-303)\right] \\ \Rightarrow \quad & 1.0011+1.2013 \times 10^{-5}(T-303) \\ & =1+1.6 \times 10^{-5}(T-303) \\ \Rightarrow 1.1 \times 10^{-3} & =(T-303) \times 10^{-5}(1.6-1.2013) \\ & =(T-303) \times 10^{-5} \times 0.3987 \\ \Rightarrow T-303 & =275 \\ \Rightarrow \quad T & =578 \mathrm{~K}=305^{\circ} \mathrm{C} \end{aligned}$

Given, initial volume, temperature and pressure be $V_1,T_1$ and $p_1$ are 1500 m$^3$, 27$^\circ$C and 4 atm.

Final volume, temperature and pressure are $V_2,T_2$ and $p_2$ are $V_2,-3^\circ$C and 2 atm

By using ideal gas equation,

$\begin{aligned} & & p V & =n R T \\ \Rightarrow & & \frac{p V}{T} & =\text { constant } \\ \Rightarrow & & \frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\ \Rightarrow & & V_2 & =\frac{p_1 V_1 T_2}{T_1 p_2} \\ & \therefore & V_2 & =\frac{4 \times 1500 \times 270}{300 \times 2}=2700 \mathrm{~m}^3 \end{aligned}$