Heat and Thermodynamics

Fractional change in time period

$ \begin{aligned} \frac{\Delta T}{T} & =\frac{1}{2} \alpha \Delta T \\ & =\frac{1}{2} \times 12 \times 10^{-6}\left(/{ }^{\circ} \mathrm{C}\right) \times 15^{\circ} \mathrm{C} \\ & =90 \times 10^{-6} \end{aligned} $

∴ The change in time per day

$ \Delta t_{\text {day }}=\left(90 \times 10^{-6}\right) \times 86400 \approx 7.8 \mathrm{~s} $

Since, the temperature increases, the length of the pendulum will increase due to thermal expansion. An increase in the length of the pendulum increases its time period, meaning the clock will tick slower and thus lose time.

The change in length due to change of temperature,

$ \begin{aligned} \Delta L & =L \alpha \Delta T \\ \Rightarrow \quad \Delta T & =\frac{\Delta L}{L \alpha}=\frac{0.5 \times 10^{-3}}{1 \times 10^{-5}} \\ & =0.5 \times 10^2=50^{\circ} \mathrm{C} \end{aligned} $

This means the temperature can vary by $\pm 50^{\circ} \mathrm{C}$ from the calibration temperature.

$ \begin{aligned} \therefore T_{\text {lower }} & =T_0-\Delta T \\ & =25^{\circ} \mathrm{C}-50^{\circ} \mathrm{C}==-25^{\circ} \mathrm{C} \\ T_{\text {upper }} & =T_0+\Delta T=25^{\circ} \mathrm{C}+50^{\circ} \mathrm{C}=75^{\circ} \mathrm{C} \end{aligned} $

For an adiabatic process,

$ \begin{aligned} & T V^{\gamma-1}=\text { constant } \\ & \begin{aligned} T_1\left[V_1\right]^{\gamma-1} & =T_2\left[V_2\right]^{\gamma-1} \\ T_1[V]^{\gamma-1} & =T_2[32 V]^{\gamma-1} \\ T_1\left[\frac{1}{32}\right]^{\gamma-1} & =T_2 \\ T_1\left[\frac{1}{32}\right]^{\frac{7}{5}-1} & =T_2 \\ {[\because \gamma} & \left.=\frac{7}{5} \text { for diatomic gas }\right] \\ T_1\left[\frac{1}{32}\right]^{\frac{2}{5}} & =T_2 \\ \frac{1}{4} & =\frac{T_2}{T_1} \end{aligned} \end{aligned} $

Efficiency of Carnot engine is given by,

$ \begin{aligned} &\eta=1-\frac{T_2}{T_1}=1-\frac{1}{4}=\frac{3}{4}=0.75\\ &\text { Percentage efficiency }=75 \% \end{aligned} $

$ \begin{aligned} &\text { Average translational kinetic energy }\\ &\begin{aligned} & E_{\text {avg }}=\frac{3}{2} K T \\ \Rightarrow & E_{\text {avg }} \propto T \\ \Rightarrow & \frac{\left(E_{\text {avg }}\right)_{\mathrm{H}_2}}{\left(E_{\text {avg }}\right)_{\mathrm{O}_2}}=\frac{T_{\mathrm{H}_2}}{T_{\mathrm{O}_2}}=\frac{T}{T}=1 \text { or } 1: 1 \end{aligned} \end{aligned} $

On combining of drop, surface area decreases, hence, heat get released,

$ \begin{aligned} & \qquad \begin{aligned} v_f & =v_i=v \\ \frac{4}{3} \pi R^3 & =n \times \frac{4}{3} \pi r^3=v \end{aligned} \\ & \Rightarrow \quad R=n^{1 / 3} r \\ & \text { Heat released, } H=S\left[A_i-A_f\right] \\ & \\ & =S\left[4 \pi r^2 \times n-4 \pi R^2\right] \\ & = \\ & =3 S\left[\frac{4 \pi r^3 \times n}{3 r}-\frac{4 \pi R^3}{3 R}\right] \\ & =3 S\left[\frac{V}{r}-\frac{V}{R}\right]=3 S V\left[\frac{1}{r}-\frac{n^{-1 / 3}}{r}\right] \\ & H=\frac{35 V}{r}\left[1-n^{-1 / 3}\right] \end{aligned} $

The faulty thermometer shows $49^{\circ} \mathrm{C}$.

To compare, let’s first change $49^{\circ} \mathrm{C}$ to Fahrenheit using the formula: $F = \left(\frac{9}{5} \times C\right) + 32$

So, $F = \left(\frac{9}{5} \times 49\right) + 32 = 88.2 + 32 = 120.2^{\circ} \mathrm{F}$

The correct Fahrenheit thermometer shows $122^{\circ} \mathrm{F}$.

The difference between the correct value and the faulty value is called the error: $122 - 120.2 = 1.8^{\circ} \mathrm{F}$.

We need to find out what this error in Fahrenheit means in Celsius.

Use the formula: $F = \frac{9}{5}C + 32$. For a difference or correction, we use only the change part: $\Delta F = \frac{9}{5} \Delta C$.

That means $\Delta C = \Delta F \times \frac{5}{9}$.

So, substitute the error: $\Delta C = 1.8 \times \frac{5}{9} = 1^{\circ} \mathrm{C}$

The faulty thermometer is $1^{\circ} \mathrm{C}$ lower than it should be, so $+1^{\circ} \mathrm{C}$ should be added to its readings.

$ \begin{aligned} & \text { Using } \lambda_{\max } T=2.9 \times 10^{-3} \\ & \qquad \begin{aligned} & \Rightarrow \quad T=\frac{2.9 \times 10^{-3}}{2.9 \times 10^{-7}}=10^4 \mathrm{~K} \\ & \text { Intensity of radiation }=\sigma T^4 \\ &= 5.67 \times 10^{-8} \times\left(10^4\right)^4 \\ &= 5.67 \times 10^8 \mathrm{~W} / \mathrm{m}^2 \end{aligned} \end{aligned} $

At constant pressure

$ \begin{aligned} \Delta U & =\frac{n f R \Delta T}{2}=\frac{5}{2} n R \Delta T \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Delta Q & =n C_p \Delta T=\frac{7}{2} n R \Delta T \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Delta W & =\Delta Q-\Delta U=n R \Delta T\left[\frac{7}{2}-\frac{5}{2}\right] \\ & =n R \Delta T \end{aligned} $

Now, ratio of

$ \Delta W: \Delta U: \Delta Q=1: \frac{5}{2}: \frac{7}{2} $

$ \text { or } \quad 2: 5: 7 $

$ \begin{aligned} & \text { } v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \\ & \Rightarrow \quad v_{\text {rms }} \propto \sqrt{T} \\ & \Rightarrow \quad \frac{\left(v_{\mathrm{rms}}\right)_2}{\left(v_{\mathrm{rms}}\right)_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{(273+527)}{(273+127)}} \\ & =\sqrt{\frac{800}{400}}=\sqrt{2} \\ & \therefore \quad\left(v_{\mathrm{rms}}\right)_2=\sqrt{2}\left(v_{\mathrm{rms}}\right)_1 \end{aligned} $

Step 1: Write what the problem says.

The Fahrenheit temperature ($F$) is $90\%$ more than the Celsius temperature ($C$). This means:

$ F = C + 0.9C = 1.9C $

Step 2: Use the formula between Celsius and Fahrenheit.

The usual formula connecting $C$ and $F$ is:

$ \frac{C}{5} = \frac{F - 32}{9} $

Step 3: Substitute what you found in Step 1 into the formula from Step 2.

Replace $F$ with $1.9C$:

$ \frac{C}{5} = \frac{1.9C - 32}{9} $

Step 4: Solve for $C$.

Multiply both sides by $9$ to get rid of the denominator:

$ 9C / 5 = 1.9C - 32 $

Now multiply both sides by $5$:

$ 9C = 5(1.9C - 32) $

$ 9C = 9.5C - 160 $

Take $9C$ from both sides:

$ 9.5C - 9C = 160 $

$ 0.5C = 160 $

So, $ C = 320 $

Step 5: Find $F$ using the value of $C$.

Use $F = 1.9C$:

$ F = 1.9 \times 320 = 608^\circ \mathrm{F} $

Use heat conduction,

$ \begin{aligned} Q & =\frac{K A \Delta T \cdot t}{d} \\ Q & =\frac{\left(10^{-2}\right) \times 0.1 \times 42 \times 9600}{2 \times 10^{-3}} \\ Q & =\frac{0.042 \times 9600}{2 \times 10^{-3}}=\frac{403.2}{2 \times 10^{-3}} \\ & =2.016 \times 10^5 \mathrm{~J} \end{aligned} $

Melted ice,

$ \begin{aligned} Q & =m_{\text {melted }} \cdot L \Rightarrow m=\frac{Q}{L} \\ m_{\text {melted }} & =\frac{2.016 \times 10^5}{336 \times 10^3}=0.6 \mathrm{~kg} \end{aligned} $

remaining ice,

$ m_{\text {remaining }}=1.5-0.6=0.9 \mathrm{~kg} $

For monoatomic gas, $C_V=\frac{3}{2} R$ and $C_p=\frac{\mathbf{5}}{\mathbf{2}} R$

For diatomic gas,

$ \begin{aligned} C_V & =\frac{5}{2} R, C_p=\frac{7}{2} R \\ \Delta U & =n C_V \Delta T \text { and } \Delta Q=n C_p \Delta T \\ \therefore \quad \frac{\Delta U}{\Delta Q} & =\frac{n C_V \Delta T}{n C_p \Delta T}=\frac{C_V}{C_p}=\frac{1}{\gamma} \Rightarrow \frac{\Delta U}{\Delta Q}=\frac{1}{\gamma} \end{aligned} $

For monoatomic gas, $\gamma_m=\frac{\frac{5}{2} R}{\frac{3}{2} R}=\frac{5}{3}$

For diatomic gas, $\gamma_d=\frac{\frac{7 R}{2}}{\frac{5 R}{2}}=\frac{7}{5}$

Given, $\Delta Q_d=\Delta Q_m=Q$

$ \begin{aligned} \therefore \quad \Delta U_m & =\frac{Q}{\gamma_m}=\frac{Q}{5}=\frac{3}{5} Q \\ \Delta U_d & =\frac{Q}{\gamma_d}=\frac{Q}{\frac{7}{5}}=\frac{5}{7} Q \end{aligned} $

$ \begin{aligned} & \therefore \quad \frac{\Delta U_m}{\Delta U_d}=\frac{\frac{3}{5} Q}{\frac{5}{7} Q}=\frac{21}{25} \\ & \therefore \Delta U_m: \Delta U_d=21: 25 \end{aligned} $

$ \text { } \begin{aligned} & V_{\text {rms }}= \sqrt{\frac{3 R T}{M}} \\ & V_{\text {rms }} \propto \sqrt{T} \\ & \Rightarrow \quad \frac{\left(V_{\text {rms }}\right)_2}{\left(V_{\text {rms }}\right)_1}=\sqrt{\frac{T_2}{T_1}} \\ &=\sqrt{\frac{273+150.5}{273+77}}=\sqrt{\frac{423.5}{350}} \\ & \Rightarrow \quad\left(V_{\text {rms }}\right)_2=\left(V_{\text {rms }}\right)_1 \times \sqrt{\frac{423.5}{350}} \\ &=50 \times \sqrt{1.21} \\ &=50 \times 1.1=55 \mathrm{~m} / \mathrm{s} \end{aligned} $

$ \begin{aligned} &\text { We know that }\\ &\begin{aligned} \Delta L & =L \propto \Delta T \\ \Rightarrow \Delta T & =\frac{\Delta L}{\Delta \alpha}=\left(\frac{\Delta L}{L}\right) \frac{1}{\alpha} \\ & =0.004 \times \frac{1}{20 \times 10^{-6}} \\ & =0.002 \times 10^5=200 \mathrm{~K} \end{aligned} \end{aligned} $

Heat released from water to bring it at $0^{\circ} \mathrm{C}$,

$ \begin{aligned} Q_1 & =m c_W \Delta T=15 \times 4200 \times 30 \\ & =1.89 \times 10^6 \mathrm{~J} \end{aligned} $

Heat released to convert water at $0^{\circ} \mathrm{C}$ to ice at $0^{\circ} \mathrm{C}$,

$ \begin{aligned} Q_2 & =m L_f=15 \times 3.34 \times 10^5 \\ & =5.01 \times 10^6 \mathrm{~J} \end{aligned} $

Total heat released

$ \begin{aligned} Q & =Q_1+Q_2 \\ & =1.89 \times 10^6+5.01 \times 10^6 \\ & =6.9 \times 10^6 \mathrm{~J} \end{aligned} $

∴ Power of refrigerator, $P=\frac{Q}{t}$

$ \begin{aligned} & =\frac{6.9 \times 10^6}{1 \times 60 \times 60} \\ & =1916.67 \mathrm{~W} \end{aligned} $

Which is nearest to 1925 W

Given, process $A B$ is isothermal Work done by the gas in isothermal process $A B$ is,

$ \begin{aligned} & =W_{A B}=n R T_1 \ln \left(\frac{V_2}{V_1}\right) \\ & =n R T_1 \ln \left(\frac{3 V_1}{V_1}\right) \\ & =n R T_1 \ln 3 \\ & =3 R T_1 \ln (3) \end{aligned} $

Compression work of process $B C$ on the gas is

$ \begin{aligned} W_{B C} & =p \Delta V=p_2\left(V_1-3 V_1\right) \\ & =-2 p_2 V_1=-\frac{2}{3} p_2\left(3 V_1\right) \\ & =-\frac{2}{3} \cdot 3 R T_1 \quad\left\{\therefore p_2\left(3 V_1\right)=3 R T_1\right\} \\ & =-2 R T_1 \end{aligned} $

Work done in process $C A=p \Delta V=0$

Total work done in cyclic process $A B C A$ is,

$ \begin{aligned} W_{A B C A} & =W_{A B}+W_{B C}+W_{C A} \\ & =3 R T_1 \ln 3-2 R T_1 \\ & =R T_1(3 \ln (3-2) \end{aligned} $

Step 1: Calculate moles of each gas

$ \begin{gathered} n_{\mathrm{O}_2}=\frac{64}{32}=2 \text{ moles} \\ n_{\mathrm{N}_2}=\frac{28}{28}=1 \text{ mole} \\ n_{\mathrm{CO}_2}=\frac{132}{44}=3 \text{ moles} \end{gathered} $

Step 2: Add up all the moles for the total initial amount

$ \begin{aligned} n_i &= n_{\mathrm{O}_2} + n_{\mathrm{N}_2} + n_{\mathrm{CO}_2} \\ &= 2 + 1 + 3 = 6 \text{ moles} \end{aligned} $

Step 3: Find out how many moles are left after removing all the oxygen gas

Only nitrogen ($\mathrm{N}_2$) and carbon dioxide ($\mathrm{CO}_2$) are left.

$ n_f = n_{\mathrm{N}_2} + n_{\mathrm{CO}_2} = 1 + 3 = 4 \text{ moles} $

Step 4: Relate pressure and moles using the formula $pV=nRT$

Since temperature and volume stay the same, the pressure is directly proportional to the number of moles. So, $\frac{p_i}{p_f} = \frac{n_i}{n_f}$

The starting pressure is $p$, so

$\frac{p}{p_f} = \frac{6}{4}$

So, the final pressure is $p_f = \frac{2}{3}p$

According to Newton's law of cooling.

$ \frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_0\right) $

For the first case

$ \begin{aligned} & \frac{60-50}{10}=K\left(\frac{60+50}{2}-T_0\right) \\ & 1=K\left(55-T_0\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

For the second case

$ \begin{aligned} & \frac{50-40}{15}=K\left(\frac{50+40}{2}-T_0\right) \\ \Rightarrow \quad & \frac{10}{15}=K\left(45-T_0\right) \\ & \frac{2}{3}=K\left(45-T_0\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Dividing Eqs. (i) by Eq. (ii), we get

$ \begin{array}{rlr} \frac{1}{2 / 3} & =\frac{55-T_0}{45-T_0} \\ \frac{3}{2} & =\frac{55-T_0}{45-T_0} \\ \Rightarrow \quad T_0 & =25^{\circ} \mathrm{C} \end{array} $

$\therefore$ From Eq. (i), $1=K(55-25)$,

$ \Rightarrow K=\frac{1}{30} $

If $t$ be the time taken to cool from $40^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$, then

$ \begin{aligned} & \frac{40-30}{t}=K\left(\frac{40+30}{2}-T_0\right) \\ & \frac{10}{t}=\left(\frac{1}{30}\right)(35-25) \\ \Rightarrow & t=30 \text { minutes } \end{aligned} $

$ \begin{aligned} &\text { According to Gay-Lussac's law, }\\ &\begin{aligned} & \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ \Rightarrow \quad & \frac{P_1}{T_1}=\frac{P_1+0.5 \% \text { of } P_1}{T_1+2.4} \\ \Rightarrow \quad & \frac{P_1}{T_1}=\frac{1.005 P_1}{T_1+2.4} \Rightarrow T_1+2.4=1.005 T_1 \\ \Rightarrow \quad & 0.005 T_1=2.4 \\ \Rightarrow \quad & T_1=\frac{2.4}{0.005}=\frac{2400}{5} \\ & =480 \mathrm{~K}=480-273^{\circ} \mathrm{C} \\ & =207^{\circ} \mathrm{C} \end{aligned} \end{aligned} $

For adiabatic process,

$ \begin{array}{ll} & T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \\ & T_1 V_1^{1.5-1}=\left(2 T_1\right) V_2^{(1.5-1)} \\ & V_1^{0.5}=2 V_2^{0.5} \\ & \Rightarrow \quad V_1=4 V_2 \Rightarrow V_2=\frac{V_1}{4} \end{array} $

$\therefore$ Percentage decrease in the volume

$ \begin{aligned} & =\frac{V_1-V_2}{V_1} \times 100 \\ & =\frac{V_1-\frac{V_1}{4}}{V_1} \times 100=75 \% \end{aligned} $

Step 1: How rms speed relates to temperature

The root mean square (rms) speed of a gas, $V_{\mathrm{rms}}$, increases as the square root of its temperature ($T$). So, $V_{\mathrm{rms}} \propto \sqrt{T}$.

Step 2: Setting up the ratio

Since $V_{\mathrm{rms}}^2 \propto T$, we can write:

$ \left[\frac{(V_{\mathrm{rms}})_1}{(V_{\mathrm{rms}})_2}\right]^2 = \frac{T_1}{T_2} $

Step 3: Substituting given values

The rms speed increases from $v$ to $\sqrt{3}v$. So:

$ \left(\frac{v}{\sqrt{3}v}\right)^2 = \frac{T_1}{T_2} $

This simplifies to $\frac{1}{3} = \frac{T_1}{T_2}$, so $T_2 = 3T_1$.

Step 4: Calculating heat required

The heat ($Q$) needed to raise the temperature is:

$ Q = nC_V\Delta T $

For 4 moles of a diatomic gas, $C_V = \frac{5}{2}R$, so:

$ Q = 4 \times \frac{5}{2} R (T_2 - T_1) $

We know $T_2 = 3T_1$, so $\Delta T = 3T_1 - T_1 = 2T_1$.

Step 5: Plugging in the numbers

We are given $Q = 83.1 \text{ kJ} = 83,100 \text{ J}$ and $R = 8.31 \text{ J mol}^{-1}\text{K}^{-1}$.

$ 83,100 = 4 \times \frac{5}{2} \times 8.31 \times 2T_1 $

$4 \times \frac{5}{2} \times 2 = 20$, so:

$ 83,100 = 20 \times 8.31 \times T_1 $

$20 \times 8.31 = 166.2$, so: $ 83,100 = 166.2 \times T_1 $

$T_1 = \frac{83,100}{166.2} = 500~\mathrm{K}$

Step 6: Converting to Celsius

$T_1$ in Celsius = $500 - 273 = 227^{\circ}\text{C}$

The heat transferred through the rod is

$ \begin{array}{rl} Q=\frac{k A \Delta T}{l} t \\& 90 \mathrm{~W} / \mathrm{mk} \times 4 \times 10^{-4} \mathrm{~m}^2 \\ Q= & \frac{\times\left(100^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}\right)}{0.2 \mathrm{~m}} \times 420 \mathrm{~s} \\ Q= & \frac{90 \times 4 \times 10^{-4} \times 100}{0.2} \times 420 \mathrm{~J} \\ Q= & \frac{36000 \times 10^{-4}}{0.2} \times 420 \mathrm{~J} \\ Q= & \frac{3.6}{0.2} \times 420 \mathrm{~J} \\ Q= & 18 \times 420 \mathrm{~J}=7560 \mathrm{~J} \end{array} $

The heat required to melt a mass $m$ of ice is given by $Q=m L_f$

$ \begin{aligned} & m=\frac{Q}{L_f} \\ & m=\frac{7560 \mathrm{~J}}{336 \times 10^3 \mathrm{~J} / \mathrm{kg}} \\ & m=\frac{7560}{336000} \mathrm{~kg} \\ & m=0.0225 \mathrm{~kg} \\ & m=22.5 \mathrm{~g} \end{aligned} $

The mass of ice melted is 22.5 g

Heat of raise temperature of ice from $-20^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$.

$ \begin{aligned} Q_1 & =m \cdot C_{\text {ice }} \cdot \Delta T \\ & =0.008 \times 2100 \times 20=336 \mathrm{~J} \end{aligned} $

Heat to melt ice at $0^{\circ} \mathrm{C}$ to water

$ Q_2=m \times L_f=0.008 \times 336000=2,688 \mathrm{~J} $

Heat to raise temperature of water from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$

$ \begin{aligned} Q_3 & =m \times C_{\text {water }} \times \Delta T \\ & =0.008 \times 4200 \times 100=3360 \mathrm{~J} \end{aligned} $

Heat to convert water at $100^{\circ} \mathrm{C}$ to steam

$ \begin{aligned} Q_4 & =M \cdot L_v=0.008 \times 2.268 \times 10^6 \\ & =18144 \mathrm{~J} \end{aligned} $

Total heat required

$ \begin{aligned} Q_{\text {Total }} & =Q_1+Q_2+Q_3+Q_4 \\ & =336+2688+3360+18144 \\ & =24,528 \mathrm{~J}(\text { or } 24.5 \mathrm{~kJ}) \end{aligned} $

1 calorie $=4.18 \mathrm{~J}$

Total heat in joules

$ Q=\frac{24528}{4.2} \approx 5840 \mathrm{cal} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} & T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \\ & T_2=T_1\left[\frac{V_1}{V_2}\right]^{\gamma-1}=600\left[\frac{V}{8 V}\right]^{\frac{4}{3}-1} \end{aligned} \end{aligned} $

$ \begin{aligned} T_2 & =600 \times \frac{1}{2}=300 \mathrm{~K} \\ W & =\frac{n R T_1}{\gamma-1}\left[1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right] \\ & =\frac{2 \times 8.3 \times 600}{\frac{4}{3}-1}\left[1-\left(\frac{1}{8}\right)^{\frac{4}{3}-1}\right] \\ & =\frac{3 \times 2 \times 8.3 \times 600}{1}\left[1-\frac{1}{2}\right] \\ & =\frac{3 \times 2 \times 8.3 \times 600}{2}=14940 \mathrm{~J} \\ & =14.94 \mathrm{~kJ} \end{aligned} $

Step 1: Recall relations

For an ideal gas,

$ C_p - C_v = R $

For a monoatomic gas,

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

So,

$ C_p = C_v + R = \frac{3}{2}R + R = \frac{5}{2}R $

Step 2: Substitute the gas constant

Given:

$ R = 8.3\;\mathrm{J\,mol^{-1}\,K^{-1}} $

Hence,

$ C_p = \frac{5}{2} \times 8.3 = 2.5 \times 8.3 = 20.75\;\mathrm{J\,mol^{-1}\,K^{-1}} $

✅ Final Answer:

$ \boxed{20.75\;\mathrm{J\,mol^{-1}\,K^{-1}}} $

Correct option: B

Temperature at which the clock loses time, $ T=38^{\circ} \mathrm{C} $

Temperature at which the clock gains

time, $ T_{2}=18^{\circ} \mathrm{C} $

$ \begin{aligned} \Delta T & =T_{1}-T_{2} \\\\ & =(38-18)^{\circ} \mathrm{C}=20^{\circ} \mathrm{C} \end{aligned} $

Time lost per day, $ \Delta t_{1}=-10.8 \mathrm{~s} $

Time gained per day, $ \Delta t_{2}=10.8 \mathrm{~s} $

The total change in time per day,

$ \Delta t=\Delta t_{1}-\Delta t_{2}=-10.8-10.8=-21.6 \mathrm{~s} $

Time period of a pendulum,

$ T=2 \pi \sqrt{\frac{L}{g}} $

The change in length $ \Delta L $ due to a temperature change $ \Delta T $ is given by

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

where, $ \alpha= $ coefficient of linear expansion

So, $ \quad \frac{\Delta L}{L}=\alpha \Delta T $

The fractional change in the time period is approximately half the fractional change in length.

$ \begin{array}{l} \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta L}{L}=\frac{1}{2} \alpha \Delta T \\\\ \frac{\Delta t_{\text {day }}}{T_{\text {day }}}=\frac{-21.6}{86400} \end{array} $

Putting value in Eq. (ii) from Eq. (i) and Eq. (iii), we get

$ \begin{aligned} \frac{-21.6}{86400} & =\frac{1}{2} \times \alpha \times 20 \\\\ \alpha & =-2.5 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} \end{aligned} $

Negative sign indicated the clock runs slower at higher temperature.

Newton's law of cooling says that,

$ \begin{array}{l} \frac{\theta_{1}-\theta_{2}}{t} \propto\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right) \\\\ \frac{\theta_{1}-\theta_{2}}{t}=k\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right) \end{array} $

where, $ \theta_{1}= $ Initial temperature, $ \theta_{2}= $ Final temperature,

$ \theta_{0}= $ Room temperature, $ t= $ time and $ k= $ constant that depends on the characteristics of the liquid and the environment.

For first cooling period

$ \begin{array}{l} \frac{368-358}{22}=k\left(\frac{368+358}{2}-\theta_{0}\right) \\\\ \frac{10}{22}=k\left(363-\theta_{0}\right) \Rightarrow k=\frac{10}{22\left(363-\theta_{0}\right)} \ldots(\mathrm{i}) \\\\ \text { For second cooling period } \\\\ \frac{358-353}{12.5}=k\left(\frac{358+353}{2}-\theta_{0}\right) \\\\ \frac{5}{12.5}=k\left(355.5-\theta_{0}\right) \Rightarrow k=\frac{5}{12.5\left(355.5-\theta_{0}\right)} \end{array} $

On comparing Eq. (i) with Eq. (ii), we get

$ \begin{aligned} \frac{10}{22\left(363-\theta_{0}\right)} & =\frac{5}{12.5\left(355.5-\theta_{0}\right)} \\\\ 25\left(355.5-\theta_{0}\right) & =22\left(363-\theta_{0}\right) \\\\ 901.5 & =3 \theta_{0} \Rightarrow \theta_{0}=300.5 \mathrm{~K} \\\\ \Rightarrow \quad \theta_{0} & =300.5-273 \\\\ \theta_{0} & =27.5^{\circ} \mathrm{C} \end{aligned} $

Given, $ U=3+1.5 \mathrm{pV} $

At constant volume, the change in internal energy is given by the change in temperature

$ d U=C_{V} d T \Rightarrow \frac{d U}{d T}=C_{V} $

We know that, $ p V=n R T $

So, $ U=3+1.5 n R T $

Putting value in Eq. (ii), we get

$ \begin{array}{l} C_{V}=\frac{d}{d T}(3+1.5 n R T) \\\\ C_{V}=1.5 n R \end{array} $

For $ C_{p} $, we use the relation

$ \begin{array}{l} C_{p}=C_{V}+n R \\\\ C_{p}=1.5 n R+n R=2.5 n R \end{array} $

The ratio of the specific heat capacities of the gas at constant volume and constant pressure is

$ \begin{array}{l} \frac{C_{V}}{C_{p}}=\frac{1.5 n R}{2.5 n R} \\\\ \frac{C_{V}}{C_{p}}=\frac{15}{25} \\\\ \frac{C_{V}}{C_{p}}=\frac{3}{5} \end{array} $

We know that,

$ p V=n R T $

Given, Initial pressure, temperature and mass of a gas are $ p, 127^{\circ} \mathrm{C} $ and 21 g respectively.

$ T_{1}=127^{\circ} \mathrm{C}=400 \mathrm{~K} $

Let $ M $ be the molar mass of the gas, then

$ p V=\left(\frac{21}{m}\right) \times R \times 400 $

After the gas leakes out,

New pressure $ =\frac{2 p}{3} $, New temperature

$ =T_{2} $

Mass of gas remaining $ =(21-5) \mathrm{g}=16 \mathrm{~g} $

Now using ideal gas equation,

$ \frac{2 p V}{3}=\left(\frac{16}{m}\right) \times R \times T_{2} $

On solving Eq. (i) and Eq. (ii), we get

$ \begin{array}{l} \frac{21}{m} \times R \times 400=\frac{16}{m} \times R \times T_{2} \times \frac{3}{2} \\\\ \frac{21 \times 400 \times 2}{16 \times 3}=T_{2} \\\\ T_{2}=350 \mathrm{~K} \\\\ T_{2}=350-273 \\\\ T_{2}=77^{\circ} \mathrm{C} \end{array} $

To determine the ratio of the masses of steam and water in equilibrium, we begin with the given data:

Mass of steam, $ m_1 = 60 \, \text{g} $

Temperature of steam, $ \theta_1 = 100^{\circ} \text{C} $

Mass of water, $ m_2 = 360 \, \text{g} $

Temperature of water, $ \theta_2 = 40^{\circ} \text{C} $

Latent heat of steam, $ L = 540 \, \text{cal/g} $

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

According to the principle of calorimetry, the heat lost by steam equals the heat gained by water:

Heat gain by water:

Using the formula $ Q = m s \Delta T $, where $ m $ is the mass, $ s $ is the specific heat, and $ \Delta T $ is the change in temperature:

$ Q = 360 \times 1 \times (100 - 40) = 21600 \, \text{cal} $

Heat lost by steam:

First, consider the latent heat: $ Q = mL = 60 \times 540 = 32400 \, \text{cal} $

Calculate the mass of steam $ x $ that would release 21600 calories of energy:

$ \begin{align*} 1 \, \text{g of steam} & = 540 \, \text{cal} \\ x \, \text{g of steam} & = 21600 \, \text{cal} \\ x & = \frac{21600}{540} = 40 \, \text{g} \end{align*} $

Thus, 40 grams of steam converted to water, releasing 21600 calories of energy.

Calculate the total mass of water after the steam condenses:

Total amount of water = initial water + condensed steam:

$ 360 \, \text{g} + 40 \, \text{g} = 400 \, \text{g} $

Calculate the remaining steam:

Steam left = initial steam - condensed steam:

$ 60 \, \text{g} - 40 \, \text{g} = 20 \, \text{g} $

Determine the ratio of steam to water:

Ratio of steam to water:

$ \frac{20}{400} = \frac{1}{20} = 1:20 $

Thus, the ratio of the masses of steam to water in equilibrium is $ 1:20 $.

Given, ratio of flow of heat (ratio) $ =\frac{5}{9} $

Ratio of radii of cylinder $ A $ and $ B=\frac{1}{2} $ We know that rate flow of heat is given as

$ \frac{d Q}{d t}=-k A \frac{d \theta}{l_{1}} $

[ $ l= $ length of conductor]

[ $ k= $ conductivity constant]

[ $ d \theta= $ temperature difference]

$ \left(\frac{d Q}{d t}\right)_{A}=-k d \theta \frac{A_{1}}{1} $

$ \begin{aligned} \frac{\left(\frac{d Q}{d t}\right)_{A}}{\left(\frac{d Q}{d t}\right)_{B}} & =\frac{A_{1}}{l_{1}} \times \frac{l_{2}}{A_{2}} \times \frac{\Delta \theta_{1}}{\Delta \theta_{2}}[k, \text { is constant }] \\\\ & =\frac{\pi r_{1}^{2}}{l_{1}} \times \frac{l_{2}}{\pi r_{2}^{2}} \times \frac{\Delta \theta_{1}}{\Delta \theta_{2}} \\\\ \frac{5}{9} & =\left(\frac{r_{1}}{r_{2}}\right)^{2} \times \frac{l_{2}}{l_{1}} \times \frac{2}{3} \\\\ \frac{5}{9} & =\frac{1}{4} \times \frac{2}{3} \times \frac{l_{2}}{l_{1}} \\\\ \frac{l_{1}}{l_{2}} & =\frac{3}{10} \\\\ l_{1}: l_{2} & =3: 10 \end{aligned} $

According to the first law of thermodynamics:

$ dQ = dU + dW $

where $ dQ $ is the heat supplied, $ dU $ is the change in internal energy, and $ dW $ is the work done by the system.

For a Monoatomic Gas:

The heat supplied $ Q_1 $ can be expressed as:

$ Q_1 = n C_V dT + W $

Here, the change in internal energy $ dU $ is given by:

$ dU = n C_V dT = n \left(\frac{3}{2}\right) R dT $

Substituting this into the equation for $ Q_1 $:

$ Q_1 = \frac{3}{2} n R dT + W $

Since $ n R dT = W $ (since $ PV = nRT $, therefore work done $ W = PV $):

$ Q_1 = \frac{3}{2} W + W $

Thus:

$ Q_1 = \frac{5W}{2} $

For a Diatomic Gas:

The heat supplied $ Q_2 $ is:

$ Q_2 = n \left(\frac{5}{2}\right) R dT + 2W $

Substituting for $ n R dT = 2W $:

$ Q_2 = \frac{5}{2} \times 2W + 2W $

This simplifies to:

$ Q_2 = 7W $

Ratio of $ Q_1 $ to $ Q_2 $:

To find the ratio $ Q_1 : Q_2 $:

$ \frac{Q_1}{Q_2} = \frac{\frac{5W}{2}}{7W} = \frac{5}{14} $

Thus, the ratio $ Q_1 : Q_2 $ is:

$ 5 : 14 $

To solve this problem, we need to find the temperature at which the root mean square (rms) speed of oxygen molecules is 75% of the rms speed of nitrogen molecules at a given temperature of $287^{\circ} \mathrm{C}$.

Given:

The rms speed of oxygen is 75% of the rms speed of nitrogen.

Temperature of nitrogen ($T_{\text{nitrogen}}$) = $287^{\circ} \mathrm{C}$.

First, convert this nitrogen temperature from Celsius to Kelvin:

$ T_{\text{nitrogen}} = 287 + 273 = 560\, \text{K} $

Formula for rms speed:

The rms speed ($v_{\text{rms}}$) is given by:

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

Where:

$R$ is the gas constant,

$T$ is the temperature in Kelvin,

$M$ is the molar mass of the gas.

For oxygen ($M_{\text{oxygen}} = 32 \, \text{g/mol}$) and nitrogen ($M_{\text{nitrogen}} = 28 \, \text{g/mol}$), we establish the following relationship by squaring both sides and canceling $R$:

$ \sqrt{\frac{3RT_{\text{oxygen}}}{32}} = \frac{3}{4} \sqrt{\frac{3RT_{\text{nitrogen}}}{28}} $

Simplifying:

Square both sides to eliminate the square roots:

$ \frac{3T_{\text{oxygen}}}{32} = \frac{9}{16} \times \frac{3T_{\text{nitrogen}}}{28} $

Cancel out the constants and solve for $T_{\text{oxygen}}$:

$ T_{\text{oxygen}} = \frac{9}{16} \times \frac{560 \times 32}{28} $

Calculate $T_{\text{oxygen}}$:

$ T_{\text{oxygen}} = 360 \, \text{K} $

Convert back to Celsius:

$ T_{\text{oxygen}} = 360 - 273 = 87^{\circ} \mathrm{C} $

Thus, the temperature at which the rms speed of oxygen achieves 75% of the rms speed of nitrogen molecules at $287^{\circ} \mathrm{C}$ is $87^{\circ} \mathrm{C}$.

When a large liquid drop splits into $n$ similar smaller drops under isothermal conditions, the following process occurs:

Volume Consistency: The total volume remains the same before and after the split. This is due to the conservation of mass principle, where the sum of the volumes of all the small drops equals the initial volume of the large drop.

Surface Area Increase: The surface area increases. The surface area $A$ of a sphere is given by $A = 4\pi r^2$, where $r$ is the radius. If a large drop with radius $R$ has a volume $V$ given by $V = \frac{4}{3}\pi R^3$, splitting into $n$ smaller drops each with radius $r$ retains the total volume as:

$ n \times \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3 $

Solving this gives:

$ n r^3 = R^3 \Rightarrow r = R \left(\frac{1}{n}\right)^{\frac{1}{3}} $

Each small drop has a surface area $4\pi r^2$, so the total surface area for $n$ small drops is:

$ n \times 4\pi r^2 = n \times 4\pi \left(R \left(\frac{1}{n}\right)^{\frac{1}{3}}\right)^2 = 4\pi R^2 \times n^{\frac{1}{3}} $

Since $n^{\frac{1}{3}} > 1$ for $n > 1$, the total surface area indeed increases when a large drop splits into smaller drops.

Energy Changes: The increase in surface area results in an increase in surface energy. According to thermodynamic principles, forming a new surface requires energy due to surface tension. Therefore, energy must be absorbed to create the additional surfaces of the new drops.

Consequently:

Option A: Incorrect, since the total volume remains constant.

Option B: Incorrect, since the total surface area increases.

Option C: Correct, energy is absorbed to increase the total surface area.

Option D: Incorrect, energy is not liberated; it is absorbed.

Thus, the correct answer is Option C: energy is absorbed.

water mass $ m_{1}=74 \mathrm{~g} $

$ \theta_{1}=70^{\circ} \mathrm{C} $

Ice mass, $ m_{2}=37 \mathrm{~g} $

$ \theta_{2}=0^{\circ} \mathrm{C} $

Ice latent heat of fusion, $ L=80 \mathrm{cal} / \mathrm{g} $

Specific heat capacity of water

$ s=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C} $

Now using principle of calorimetry

Heat lost by water is equal to heat gain by ice. Let $ \theta $ be the common temperature.

$ \begin{aligned} m_{1} s\left(\theta_{1}-\theta\right) & =m_{2} L+m_{2} s\left(\theta-\theta_{2}\right) \\\\ 74 \times 1 \times[70-\theta] & =37[80+1(\theta-0)] \\\\ 2[70-\theta] & =[80+\theta] \\\\ 140-2 \theta & =80+\theta \\\\ 140-80 & =3 \theta \Rightarrow 60=3 \theta \\\\ \theta & =20^{\circ} \mathrm{C} \end{aligned} $

Given,

Metal plate thickness,

$ d x=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m} $

Area of plate, $ A=5 \mathrm{~cm}^{2}=5 \times 10^{-4} \mathrm{~m}^{2} $

Temperature difference,

$ d T=14^{\circ} \mathrm{C}=14 \mathrm{~K} $

Rate of thermal flow $ =42 \mathrm{~J} $

We know that equation for rate of thermal flow is

$ \begin{aligned} \frac{d Q}{d t} & =-K A \frac{d T}{d x} \\\\ \frac{d Q}{d T} & =-42 \mathrm{~J} \quad[\text { heat flows out }] \\\\ -42 & =\frac{-K \times 5 \times 10^{-4} \times 14}{5 \times 10^{-3}} \\\\ 42 & =1.4 K \\\\ K & =30 \mathrm{~W} / \mathrm{m} / \mathrm{K} \end{aligned} $

Given, ratio of $ \frac{C_{p}}{C_{V}} $ or $ \gamma=1.5 $

For adiabatic process,

Let initial pressure $ p_{a} $, Final pressure $ =p_{1} $ Initial volume $ =V $, Final volume $ =2 V $

Using adiabatic equation

$ \begin{aligned} p V^{\gamma} & =\text { constant } \\\\ p_{a} V_{1}^{\gamma} & =p_{1} V_{2}^{\gamma} \\\\ \text { (Initial) } & \text { (Final) } \\\\ \Rightarrow \quad & p_{a} \end{aligned}=p_{1}\left[\frac{2 V}{V}\right]^{\gamma} \text {, } $

For isothermal process, final pressure $ =p_{2} $ Using isothermal equation

$ \begin{aligned} p V & =\text { constant } \\\\ p_{\text {iso }} V_{1} & =p_{2} V_{2} \\\\ \text { (Initial) } & =(\text { Final) } \end{aligned} \quad \begin{aligned} p_{\text {iso }} & =p_{2}\left[\frac{2 V}{V}\right] \\\\ \Rightarrow \quad p_{2} & =\frac{p_{\text {iso }}}{2} \end{aligned} $

According to question,

$ \begin{aligned} p_{1} & =p_{2} \\\\ \frac{p_{a}}{[2]^{1.5}} & =\frac{p_{\text {iso }}}{2} \\\\ \frac{p_{a}}{p_{\text {iso }}} & =\frac{2 \sqrt{2}}{2} \quad\left[\text { as } a^{1.5}=a \cdot \sqrt{a}\right] \\\\ p_{a}: p_{\text {iso }} & =\sqrt{2}: 1 \end{aligned} $

The average translational kinetic energy (per molecule) of any ideal gas (diatomic or monoatomic or polyatomic) is always $ \frac{3}{2} k_{B} T $.

It depends only on temperature and is independent of nature of gas.

Hence, the ratio of two gases is $ 1: 1 $.

Given, mass of ice at $-20^{\circ} \mathrm{C}$,

$ m_{\mathrm{ice}}=54 \mathrm{~g} $

Mass of steam at $100^{\circ} \mathrm{C}, m_{\text {meam }}=25 \mathrm{~g}$

Specific heat capacity of ice,

$ s_{\text {iee }}=0.5 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C} $

Heat required to bring ice at $0^{\circ} \mathrm{C}$

$ \begin{aligned} & =m_{\text {ce }} s_{\text {cle }} \Delta T \\\\ q_1 & =54 \times 0.5 \times(20)=540 \mathrm{cal} \end{aligned} $

Heat required to melt ice at $0^{\circ} \mathrm{C}$

$ \begin{aligned} & =m_{\mathrm{ice}} \times L_{\mathrm{ice}} \\\\ & q_2=54 \times 80=4320 \mathrm{cal} \end{aligned} $

Condense steam at $100^{\circ} \mathrm{C}$ to water at $100^{\circ} \mathrm{C}$.

$ \begin{aligned} q_3 & =m_{\text {steam }} \times L_{\text {steam }}=25 \times 540 \\\\ & =13500 \mathrm{cal} \end{aligned} $

To cold water from $100^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$,

$ \begin{aligned} q_4 & =m_{\text {water }} \times s_{\text {water }} \times \Delta T \\\\ & =25 \times 1 \times 100=2500 \mathrm{cal} \end{aligned} $

$\therefore$ Total heat gained by ice,

$ Q_{\text {gain }}=q_1+q_2=4860 \mathrm{cal} $

Heat lost by steam, $Q_{\text {lott }}=q_3=13500 \mathrm{cal}$ Heat gained by ice to melt will be provided by steam i.e, $\frac{4860}{540}=9 \mathrm{~g}$

Thus, only 9 g steam is needed to completely melt the ice,

Hence, $(54+9) \mathrm{g}$ of water will be present along with $(25-9) \mathrm{g}$ of steam at $100^{\circ} \mathrm{C}$.

Lets calculate the temperature of mixture.

Remaining heat of steam;

$ \begin{aligned} Q & =16 \times 540 \\\\ & =8640 \mathrm{cal} \end{aligned} $

Heat required to raise the tempature of 63 g upto $T^{\circ}$ will be

$ \begin{array}{rlrl} Q & =m s_{\text {water }} \times(\Delta T) \\\\ 8640 & =63 \times 1(\Delta T) \\\\ \Rightarrow \quad \Delta T & >100^{\circ} \mathrm{C} \end{array} $

Hence, not all steam will be converted into water. The steam required to raise 63 g water upto $100^{\circ} \mathrm{C}$ will be,

$ m=\frac{6300}{540}=11.66 \mathrm{~g} $

$\therefore$ Total steam used $=11.66+9$

$ \approx 20 \mathrm{~g} $

Hence 20 g of steam will be used to bring 54 g of ice at $-20^{\circ} \mathrm{C}$ upto $100^{\circ} \mathrm{C}$ water.

Total water $=54+20 \approx 74 \mathrm{~g}$

and steam $=5 \mathrm{~g}$

Which is close to option (b).

Total surface area of solid sphere

$ A_1=4 \pi r^2 $

Total surface area of solid hemisphere,

$ A_2=3 \pi r^2 $

As, we know that, $E \propto A$

$ \therefore \quad \frac{E_2}{E_1}=\frac{A_2}{A_1}=\frac{3}{4} $

For diatomic gas, $C_p=\frac{7}{2} R$

$ \begin{aligned} C_v & =\frac{5}{2} R \\\\ \therefore \quad d U & =n C_v d T \end{aligned} $

$ \begin{aligned} d Q & =n C_p d T \\\\ d W & =p \times d V \\\\ & =n R d T \\\\ \therefore d W: d U: d Q & =n R d T: n \frac{5}{2} R d T: n \frac{7}{2} R d T \\\\ & =1: \frac{5}{2}: \frac{7}{2} \\\\ & =2: 5: 7 \end{aligned} $

We know that, $v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$

According to question, $v_1=\sqrt{\frac{3 R(300)}{M}}$

$ v_2=\sqrt{\frac{3 R(432)}{M}} $

Thus, $\frac{v_2-v_1}{v_1}=\frac{\sqrt{432}-\sqrt{300}}{\sqrt{300}}=\frac{3.46}{17.32}$

$ \begin{aligned} \% \text { increase } & =\frac{v_2-v_1}{v_1} \times 100 \% \\\\ & =\frac{3.46}{17.32} \times 100 \% \\\\ & =19.99 \% \\\\ & \approx 20 \% \end{aligned} $

The relationship between the two scales,

$ F=\frac{9}{5} C+32 \quad \ldots \text { (i) } $

If the temperature is twice the celsius temperature, we can write

$ F=2 C \quad \ldots \text { (ii) } $

Now, $\frac{9}{5} \times C+32=2 C$

$ \begin{aligned} & \frac{9}{5} \times C+32=\frac{10}{5} C \\\\ & 32=\frac{10}{5} C-\frac{9}{5} C \Rightarrow 32=\frac{1}{5} C \\\\ & C=160 \quad \ldots \text { (iii) } \end{aligned} $

Fahrenheit temperature,

$ \begin{aligned} F & =\frac{9}{5} \times 160+32 \\\\ \quad F & =288+32 \\\\ \Rightarrow \quad F & =320 \end{aligned} $

Therefore, the temperature on the fahrenheit scale that is twice the temperature of $160^{\circ} \mathrm{C}$ on the celsius scale is $320^{\circ} \mathrm{F}$.

For liquid $A$ and $B$ mixed to reach a final temperature of $20^{\circ} \mathrm{C}$, the equation based on the heat exchanged is

$ m \times C_B \times(24-20)=m \times C_A \times(20-15) $

Simplifying, we get

$ \begin{aligned} 4 m \times C_B & =5 m \times C_A \\\\ \frac{C_B}{C_A} & =\frac{5}{4} \end{aligned} $

For liquid $B$ and $C$ mixed to reach a final temperature of $26^{\circ} \mathrm{C}$,

$ m \times C_C \times(30-26)=m \times C_B \times(26-24) $

we get,

$ \begin{aligned} 4 m \times C_C & =2 m \times C_B \\\\ \frac{C_C}{C_B} & =\frac{1}{2} \Rightarrow \frac{C_B}{C_A} \times \frac{C_C}{C_B}=\frac{5}{4} \times \frac{1}{2}=\frac{5}{8} \end{aligned} $

Therefore,

$ \begin{aligned} & C_A: C_B: C_C=1: \frac{5}{4}: \frac{5}{8} \\\\ & C_A: C_B: C_C=8: 10: 5 \end{aligned} $

The efficiency of a reversible heat engine is given by the ratio of work output to heat input, which can be also be expressed in terms of the temperatures of the hot $\left(T_H\right)$ and cold $\left(T_C\right)$ reservoirs,

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

Given, efficiency is $\mathbf{5 0 \%}$

$ 0.5=1-\frac{T_C}{T_H} $

Solving for $\frac{T_C}{T_H}$, we get

$ \frac{T_C}{T_H}=0.5 $

Coefficient of performance (COP),

$ \mathrm{COP}=\frac{Q_C}{W}=\frac{T_C}{T_B-T_C} $

For a reversible process, the (COP) can also be related to the temperatures of the hot and cold reservoirs,

$ \begin{aligned} \mathrm{COP} & =\frac{0.5 \times T_H}{T_H-0.5 \times T_H} \\\\ & =\frac{0.5}{1-0.5}=\frac{0.5}{0.5}=1 \end{aligned} $

Therefore, the coefficient of performance of the refrigerator working between the same two temperature but reverse direction is 1.

Given, $n=4$ moles

Temperature,

$ T=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K} $

We know that,

$ U=\frac{5}{2} \times n \times R \times T \quad \ldots \text { (i) } $

Putting above values in Eq. (i),

$ U=\frac{5}{2} \times 4 \times 8.314 \times 300 \mathrm{~K} $

$\therefore$ Universal gas constant,

$ \begin{aligned} R & =8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\\\ U & =5 \times 2 \times 8.31 \times 300 \\\\ & =10 \times 8.31 \times 300 \\\\ & =3000 \times 8.31 \\\\ U & =24930 \mathrm{~J}=24.930 \mathrm{~J} \end{aligned} $

Thus, $U=24.93 \mathrm{~kJ}$

Given,

Steam temperature $=100^{\circ} \mathrm{C}$

Initial water mass $=150 \mathrm{~g}$

Initial temperature of water $=20^{\circ} \mathrm{C}$

Final temperature $=40^{\circ} \mathrm{C}$

Specific heat of water $=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}$

Latent heat of steam $=540 \mathrm{cal} / \mathrm{g}$

Using principle of calorimetry.

Heat lost by steam = Heat gained by water

$ \begin{aligned} m L_{\text {vopourisation }}+m S_w \Delta T & =m_w S_w \Delta T \\ m \times 540+m \times 1 & \times(100-40) \\ & =150 \times 1 \times(40-20) \\ 540 m+60 m & =3000 \\ 600 m & =3000 \\ m & =5 \mathrm{~g} \end{aligned} $

Hence, total mass $=150+5=155 \mathrm{~g}$

Given,

Diameter of wooden wheel $=5.012 \mathrm{~m}$ Diameter of circular iron frame $=5 \mathrm{~m}$ Temperature, $T=27^{\circ} \mathrm{C}$

Length increases thermally is given by

$ \begin{aligned} \Delta L & =L_0 \alpha \Delta T \\ \text { where } \Delta L & =L_1-L_0 \\ \Delta T & =T_2-T_1 \end{aligned} $

$L_0=$ Initial diameter,$L_1=$ Final diameter [Diameter of circular frame, Diameter of wooden wheel]

$ \Delta T=\frac{\Delta L}{\alpha L_0} $

[ $\alpha$-Linear expansion coefficient $\left.=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}\right]$

$ \begin{aligned} & T_2-T_1=\frac{L_1-L_0}{\alpha L_0} \\ & \left.T_2-27^{\circ} \mathrm{C}=\frac{5.012-5.000}{5 \times 1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}}\right] \\ & \Rightarrow \quad T_2=\frac{0.012}{5 \times 1.2 \times 10^{-5}}+27^{\circ} \mathrm{C} \\ & \quad=227^{\circ} \mathrm{C} \end{aligned} $

Hence, the required temperature is $227^{\circ} \mathrm{C}$.

Given,

Number of moles, $n=2$

$ \gamma=\frac{4}{3} \quad \text { (for triatomic gas) } $

$C_V$ for triatomic gas $=3 R$

Initial volume $=V_i$

Final volume $=8 V_i=V_f$

In adiabatic process work done is given by

$ \Delta U=n C_V \Delta T $

We have to find $\Delta T$ first, for that $p V^{\gamma-1}=$ constant (adiabatic condition)

$ \begin{aligned} & \frac{T_2}{T_1}=\left(\frac{V_f}{V_i}\right)^{\gamma-1} \\ & \quad\left[T_1=327^{\circ} \mathrm{C}=600 \mathrm{~K} \text { given }\right] \\ & T_2=T_1\left(\frac{1}{8}\right)^{\frac{4}{3}-1} \Rightarrow T_2=\frac{600}{2}=300 \mathrm{~K} \end{aligned} $

Now work done is $\Delta U=n C_V \Delta T$

$ =2 \times 3 R \times 300=1800 R $

Work done in isochoric process is zero because volume remains constant after above process.

Hence, work done is $1800 R$.

Given,

Initial temperature $T_1=27^{\circ} \mathrm{C}$

Final temperature $T_2=159^{\circ} \mathrm{C}$

RMS speed of gas is given by

$ \begin{gathered} v_{\text {rms }}=\sqrt{\frac{3 K T}{m}} \\ v_{\text {rms }} \propto \sqrt{T} \\ \text { for } T=27^{\circ} \mathrm{C}=300 \mathrm{~K} \\ v_{\text {rms }} \propto \sqrt{300}=17.32 \mathrm{~m} / \mathrm{s} \\ v_{\text {rms }} \propto \sqrt{432}=20.78 \mathrm{~m} / \mathrm{s} \end{gathered} $

Percentage increase

$ \frac{\Delta v_{\mathrm{rms}}}{v_{\mathrm{rms}}} \times 100 \Rightarrow \frac{20.78-17.32}{17.32} \times 100=20 \% $

Hence, option (d) is the correct option.