Heat and Thermodynamics

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

Given Values:

Let $C_1$ be the specific heat of the metal and $C_2$ be the specific heat of water.

The ratio of their specific heats is $\dfrac{C_1}{C_2} = \dfrac{1}{10}$, so $C_2 = 10C_1$.

A piece of metal with mass $100$ g gets some heat and its temperature increases by $20^\circ \mathrm{C}$.

The same amount of heat is then given to $20$ g of water. We must find how much the temperature of water changes ($\Delta T$).

Step 1: Find Heat Given to Metal

The heat required to change temperature is given by $\Delta Q = m C \Delta T$.

Here: $m = 100$ g, $C = C_1$, and $\Delta T = 20$.

So, $\Delta Q = 100 C_1 \times 20$.

Step 2: Set Heat Given to Water Equal to Heat Given to Metal

For water: $m = 20$ g, $C = C_2$, and temperature change is $\Delta T$.

So, $\Delta Q = 20 \times C_2 \times \Delta T$.

Since both heats ($\Delta Q$) are the same:

$100 C_1 \times 20 = 20 \times C_2 \times \Delta T$

Step 3: Substitute the Value of $C_2$

We know $C_2 = 10C_1$. Substitute this value:

$100 C_1 \times 20 = 20 \times 10C_1 \times \Delta T$

Step 4: Simplify to Find $\Delta T$

$2000 C_1 = 200 C_1 \times \Delta T$

Divide both sides by $200 C_1$:

$\Delta T = \dfrac{2000 C_1}{200 C_1} = 10^\circ \mathrm{C}$

So, the temperature of the water increases by $10^\circ \mathrm{C}$.

$ \text { } \begin{aligned} \eta_A & =1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}=\frac{\Delta T}{T_1} \\ \eta_B & =1-\frac{T_2^{\prime}}{T_1^{\prime}}=\frac{T_1^{\prime}-T_2^{\prime}}{T_1^{\prime}}=\frac{\Delta T}{T_1^{\prime}} \\ \therefore \quad \frac{\eta_A}{\eta_B} & =\frac{\frac{\Delta T}{T_1}}{\frac{\Delta T}{T_1^{\prime}}} \Rightarrow 1.25=\frac{T_1^{\prime}}{T_1} \\ \Rightarrow \quad \frac{T_1^{\prime}}{T_1} & =1.25=\frac{125}{100} \\ & \frac{T_1^{\prime}}{T_1}=\frac{5}{4} \Rightarrow \frac{T_1}{T_1^{\prime}}=\frac{4}{5} \end{aligned} $

$ \begin{aligned} &\text { Work done, }\\ &\begin{aligned} W & =P \Delta V=P\left(V_2-V_1\right) \\ & =5 \times 10^5(2.5-1)=7.5 \times 10^5 \mathrm{~J} \end{aligned} \end{aligned} $

According to first law of thermodynamics

$ \begin{aligned} \Delta U & =Q-W=1000 \times 10^3-7.5 \times 10^5 \\ & =2.5 \times 10^5=250 \times 10^3 \mathrm{~J}=250 \mathrm{~kJ} \end{aligned} $

$ \begin{aligned} &\text { For adiabatic process }\\ &\begin{aligned} & p V^\gamma=\text { constant } \\ \Rightarrow & \ln p+\gamma \ln (V)=\text { constant } \\ \Rightarrow \quad & \frac{d p}{p}+\gamma \frac{d V}{V}=0 \Rightarrow \frac{d p}{p}=-\gamma \frac{d V}{V} \\ \Rightarrow \quad & \frac{\Delta p}{p}=-\gamma \frac{\Delta V}{V} \\ & =-1.4 \times 0.005=-0.007 \\ \therefore & \frac{\Delta p}{P} \times 100 \%=-0.007 \times 100 \%=-0.7 \% \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \\ & \Rightarrow \quad V_{\mathrm{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}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} &\text { } \begin{aligned}Since, \left(V_{\mathrm{rms}}\right)_2 & =\left(V_{\mathrm{rms}}\right)_1+25 \% \text { of }\left(V_{\mathrm{rms}}\right)_1 \\ & =\left(V_{\mathrm{rms}}\right)_1+\frac{1}{4}\left(V_{\mathrm{rms}}\right)_1 \\ & =\frac{5}{4}\left(V_{\mathrm{rms}}\right)_1 \end{aligned}\\ &\therefore \text { From equation (i), we get }\\ &\begin{array}{ll} & \frac{\frac{5}{4}\left(V_{\mathrm{rms}}\right)_1}{\left(V_{\mathrm{rms}}\right)_1}=\sqrt{\frac{T_2}{T_1}} \Rightarrow \frac{T_2}{T_1}=\frac{25}{16} \\ \therefore & \frac{T_2-T_1}{T_1} \times 100 \%=\left(\frac{T_2}{T_1}-1\right) \times 100 \% \\ & =\left(\frac{25}{16}-1\right) \times 100 \% \\ & =\frac{9}{16} \times 100 \%=56.25 \% \end{array} \end{aligned} $

$\frac{K_{\mathrm{Cu}}}{K_{\mathrm{Br}}}=\frac{4}{1}=4 \Rightarrow K_{\mathrm{Cu}}=4 K_{\mathrm{Br}}$

According to given condition, Heat flow through copper $=$ Heat flow

$ \begin{array}{ll} \Rightarrow & \frac{K_{\mathrm{Cu}} A\left(100-T_0\right)}{L}=\frac{K_{\mathrm{Br}}\left(T_0-0\right) A}{L} \\ \Rightarrow & 4 K_{\mathrm{Br}}\left(100-T_0\right)=K_{\mathrm{Br}} T_0 \\ \Rightarrow & 400-4 T_0=T_0 \\ \Rightarrow & 5 T_0=400 \\ \Rightarrow & T_0=80^{\circ} \mathrm{C} \end{array} $

Initial efficiency of Carnot engine.

$ \begin{aligned} \eta & =1-\frac{T_2}{T_1} \\ \frac{1}{3} & =1-\frac{T_2}{T_1} \\ \Rightarrow \quad \frac{T_2}{T_1} & =1-\frac{1}{3}=\frac{2}{3} \Rightarrow T_1=\frac{3}{2} T_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

New efficiency,

$ \begin{gathered} \eta^{\prime}=1-\frac{T_2^{\prime}}{T_1^{\prime}} \\ \frac{3}{16}=1-\frac{T_2+25}{T_1-50} \\ \Rightarrow \frac{T_2+25}{T_1-50}=1-\frac{3}{16}=\frac{13}{16} \\ \Rightarrow 16 T_2+400=13 T_1-650 \\ \Rightarrow 16 T_2+1050=13 T_1 \end{gathered} $

$ \begin{aligned} & 16 T_2+1050=13\left(\frac{3}{2} T_2\right)[\text { From eqn }(\mathrm{i})] \\ & \Rightarrow 32 T_2+2100=39 T_2 \\ & \Rightarrow 7 T_2=2100 \\ & T_2=\frac{2100}{7}=300 \mathrm{~K} \end{aligned} $

Since, $p V=n R T$

at constant pressure,

$ \begin{array}{rlr} V \propto T & \\ \frac{V_1}{V_2} & =\frac{T_1}{T_2} \Rightarrow \frac{V}{3 V}=\frac{T_1}{T_2} \\ \Rightarrow \quad T_2 & =3 T_1 \\ \therefore \quad \Delta T & =T_2-T_1=3 T_1-T_1=2 T_1 \\ \Delta T & =2 T \\ \Delta U & =n C_V \Delta T \\ & =n C_V(2 T) \\ & =n\left(\frac{R}{\gamma-1}\right) 2 T \quad\left[\because C_V=\frac{R}{\gamma-1}\right] \\ & =\frac{2 n R T}{\gamma-1}=\frac{2 p V}{\gamma-1} \quad[\because n R T=p V] \end{array} $

For adiabatic process

$ \begin{aligned} p_1 V_1^\gamma & =p_2 V_2^\gamma \\ \frac{p_2}{p_1} & =\left(\frac{V_1}{V_2}\right)^\gamma=\left(\frac{V_1}{8 V_1}\right)^{5 / 3} \\ & =\left[\left(\frac{1}{2}\right)^3\right]^{5 / 3}=\left(\frac{1}{2}\right)^5=\frac{1}{32} \\ p_2 & =\frac{1}{32} \times p_1=\frac{1}{32} \times 100 \times 10^3 \mathrm{~Pa} \end{aligned} $

Now, work done in adiabatic process is,

$ W=\frac{p_1 V_1-p_2 V_2}{\gamma-1} $

$\Rightarrow 180=\frac{100 \times 10^3 \times V_1-\frac{1}{32} \times 100 \times 10^3\left(8 V_1\right)}{\frac{5}{3}-1} $

$ \begin{aligned} \Rightarrow 180 \times \frac{2}{3} & =100 \times 10^3 \times V_1\left(1-\frac{8}{32}\right) \\ 180 \times \frac{2}{3} & =\frac{3}{4} \times 100 \times 10^3 \times V_1 \\ \Rightarrow \quad V_1 & =\frac{40 \times 4}{100 \times 10^3} \mathrm{~m}^3 \\ & =\frac{40 \times 4}{100 \times 10^3} \times(100)^3 \\ & =1600 \mathrm{~cm}^3 \end{aligned} $

We know that internal energy

$ U=\frac{f}{2} n R T $

∴ Total internal energy of mixture

$ \begin{aligned} & =U_{\mathrm{O}_2}+U_{\mathrm{Ar}} \\ & =\frac{5}{2} \times 3 \times R T+\frac{3}{2} \times 4 \times R T \\ & =13.5 R T \end{aligned} $

Newton's Law of Cooling:

Newton's law of cooling says that the rate at which something cools down depends on the difference between its temperature and the temperature of its surroundings.

The formula for Newton's law of cooling is:

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

Step 1: Use the formula for the first 10 minutes

The body cools from $62^\circ \mathrm{C}$ to $50^\circ \mathrm{C}$ in 10 minutes.

• Here, $T_1 = 62$, $T_2 = 50$, and $t = 10$.
• Plug these into the formula:

$ \frac{62-50}{10}=K\left(\frac{62+50}{2}-T_0\right) $

This simplifies to:

$ 1.2=K(56-T_0) \quad ...(i) $

where $56$ comes from $(62+50)/2$.

Step 2: Use the formula for the next 10 minutes

The body cools from $50^\circ \mathrm{C}$ to $42^\circ \mathrm{C}$ in the next 10 minutes.

• Now, $T_1 = 50$, $T_2 = 42$, and $t = 10$.
• Plug these values in:

$ \frac{50-42}{10}=K\left(\frac{50+42}{2}-T_0\right) $

This simplifies to:

$ 0.8=K(46-T_0) \quad ...(ii) $

where $46$ comes from $(50+42)/2$.

Step 3: Set up a ratio to find $T_0$

Divide equation (i) by equation (ii) (the $K$ cancels out):

$ \frac{1.2}{0.8} = \frac{K(56-T_0)}{K(46-T_0)} $

So,

$ \frac{1.2}{0.8} = \frac{56 - T_0}{46 - T_0} $

Step 4: Solve for $T_0$

$\frac{1.2}{0.8} = 1.5$

Set up the equation:

$1.5 = \frac{56 - T_0}{46 - T_0}$

Multiply both sides by $(46 - T_0)$:

$1.5 (46 - T_0) = 56 - T_0$

$69 - 1.5 T_0 = 56 - T_0$

$69 - 56 = 1.5 T_0 - T_0$

$13 = 0.5 T_0$

$T_0 = 26^\circ \mathrm{C}$

Final Answer:

The temperature of the surroundings is $26^\circ \mathrm{C}$.

The formula for specific heat at constant volume is $C_V = \frac{f}{2} R$, where $f$ is the number of degrees of freedom and $R$ is the universal gas constant.

For a monoatomic gas, $f = 3$.

So, $C_V = \frac{3}{2} R$.

This means $\frac{R}{C_V} = \frac{R}{\frac{3}{2}R} = \frac{2}{3} = 0.67$.

Therefore, the gas is monoatomic because it matches the given ratio of 0.67.

Internal energy,

$ U=\frac{f}{2} n R T $

For monoatomic gas, $f=3$

$ \begin{aligned} U & =\frac{3}{2} n R T=\frac{3}{2} \times 4 \times R \times(273+77) \\ & =6 R \times 350=2100 R \end{aligned} $

For monoatomic gas, $\gamma=\frac{5}{3}$

For adiabatic process,

$ \begin{aligned} & T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \\ \Rightarrow \quad & T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1} \\ & =4 T_1 \\ \therefore \quad W & =\frac{n R\left(T_2-T_1\right)}{\gamma-1} \\ & =\frac{0.25 R\left(4 T_1-T_1\right)}{\frac{5}{2}-1} \\ & \\ & \quad\left[\because n=\frac{5.6}{22.4}=0.25 \mathrm{~mol}\right] \\ & \quad= 0.25 R \times \frac{3}{2} \times 3 T_1 \\ & \quad= 4.5 \times 0.25 R \times 273 \\ & \simeq 307.29 R \\ & \simeq 307 R \end{aligned} $

$ \begin{aligned} \text { } v_{\mathrm{rms}} & =\sqrt{\frac{3 R T}{M}} \\ \Rightarrow \quad 2000 & =\sqrt{\frac{3 \times 8.31 \times 322}{M}} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 4000000=\frac{8027.5}{M} \\ & \Rightarrow \quad M=\frac{8027.5}{4000000} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& =0.002007 \mathrm{~kg} / \mathrm{mole} \\ & =2.007 \mathrm{~g} / \mathrm{mole} \approx 2 \mathrm{~g} / \mathrm{mole} \end{aligned}\\ &\text { Therefore, gas is hydrogen. } \end{aligned} $

Let $m=$ mass of steam condensed. Then, $m$ grams of steam forms $m$ grams of water at $100^{\circ} \mathrm{C}$ which further losses heat and comes at $90^{\circ} \mathrm{C}$.

Final mixture is water at $90^{\circ} \mathrm{C}$

Now using Heat lost = Heat gained, we gel

Heat lost by steam in condensation + Heat lost by water formed at $100^{\circ} \mathrm{C}$

$=$ Heat gained by water at $24^{\circ} \mathrm{C}$

$ \begin{aligned} & \Rightarrow m L+m c(100-90)=100(c)(90-24) \\ & \Rightarrow m(540+1 \times 10)=100 \times 1 \times 66 \\ & \Rightarrow m=\frac{100 \times 66}{550}=\frac{10 \times 6}{5}=12 \mathrm{~g} \end{aligned} $

Heat supplied at constant pressure is used to do work and to raise temperature of the gas.

$ \begin{aligned} \therefore & & \Delta Q & =\Delta W+\Delta U \\ \Rightarrow & & 80 & =20+\Delta U \\ \Rightarrow & & \Delta U & =60 \mathrm{~J} \end{aligned} $

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

and $\Delta U=n C_V \Delta T$

$\therefore$ Ratio $\frac{C_p}{C_V}=\frac{\Delta Q}{\Delta U}=\frac{80}{60}=\frac{4}{3}$

Step 1: Understand the COP formula

The coefficient of performance (COP) for a refrigerator shows how well it works. The formula is:

$ \mathrm{COP} = \frac{Q_H}{Q_H - Q_C} $

Step 2: Identify what the symbols mean

$Q_H$ stands for the heat given to the room (hot side). This is the heat released.

$Q_C$ is the heat taken from inside the fridge (cold side). This is the heat extracted, which is 250 J per cycle here.

Step 3: Set up the equation with the given numbers

We are told the COP is 5 and $Q_C$ is 250 J. So, we use:

$ 5 = \frac{Q_H}{Q_H - 250} $

Step 4: Solve for $Q_H$

Multiply both sides by $(Q_H - 250)$:

$ 5(Q_H - 250) = Q_H $

Expand:

$ 5Q_H - 1250 = Q_H $

Bring $Q_H$ to one side:

$ 5Q_H - Q_H = 1250 $

$ 4Q_H = 1250 $

Divide both sides by 4:

$ Q_H = \frac{1250}{4} $

$ Q_H = 312.5~\mathrm{J} $

Step 5: Answer

The refrigerator releases about 312.5 J of heat to the room per cycle.

Using $p V=n R T$

we get, $\quad p=\frac{n R T}{V}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Here, $n=n_1+n_2+n_3$

$ \begin{gathered} =2+5+3=10 \\ V=16.62 \mathrm{~m}^3, T=0^{\circ} \mathrm{C}=273 \mathrm{~K} \end{gathered} $

Substituting in (i) we get,

$ \begin{aligned} p & =\frac{10 \times 8.31 \times 273}{16.62} \\ & =1365 \mathrm{~Pa} \end{aligned} $

Given:

• Mass of water = $ m $

• Temperature of water = $ \theta^{\circ} \mathrm{C} $

• Mass of ice = $ M $

• Ice is at its melting point ( $ 0^{\circ} \mathrm{C} $ )

• $ s $ = specific heat capacity of water

• $ L $ = latent heat of fusion of ice.

Concept:

When water (at temperature $\theta$) is mixed with ice (at $0^{\circ}$), heat will flow from the warmer water to the ice.

• The water cools down from $ \theta $ to $ 0^{\circ} \mathrm{C}$.

• The ice melts, using that heat.

No heat is lost to surroundings.

Step 1: Heat lost by water cooling to $0^{\circ} \mathrm{C}$

$ Q_{\text{lost}} = m s \theta $

Step 2: Heat gained by ice (for melting)

If $ x $ is the mass of ice melted, then:

$ Q_{\text{gained}} = x L $

Step 3: Equating heat lost and gained

$ m s \theta = x L $

Step 4: Solve for $ x $

$ x = \frac{m s \theta}{L} $

Step 5: Check if $ M $ (the large mass of ice) matters

Since $ M $ is very large, its temperature doesn’t rise; it only partly melts.

Hence $ M $ appears only to indicate that the ice remains at 0°C.

✅ Final Answer:

$ \boxed{x = \frac{m s \theta}{L}} $

Correct Option: D

The efficiency formula for a Carnot engine is:

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

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

The problem says the source is 25% hotter than the sink. That means:

$ T_1 = T_2 + 0.25 \times T_2 = 1.25 T_2 $

Now, plug $T_1 = 1.25T_2$ into the efficiency formula:

$ \eta = 1 - \frac{T_2}{1.25T_2} $

$ \eta = 1 - \frac{1}{1.25} $

$ \frac{1}{1.25} = \frac{4}{5} $

So,

$ \eta = 1 - \frac{4}{5} $

$ \eta = \frac{1}{5} $

To change this to a percentage, multiply by $100\%$:

$ \eta = \frac{1}{5} \times 100\% = 20\% $

So, the engine’s efficiency is $20\%$.

Step 1: Use the ideal gas law

We know that $pV = nRT$ describes how pressure, volume, and temperature are related for an ideal gas.

Step 2: Set up the formula for work at constant pressure

When pressure is constant, the work ($W$) done by the gas is:

$W = p \Delta V$

From the ideal gas law, we can find that:

$p \Delta V = nR \Delta T$

This means the work done is:

$W = nR \Delta T$

Step 3: Insert the given values

Number of moles, $n = 6$

Change in temperature, $\Delta T = 20^\circ \mathrm{C}$. (This is the same as $20~\mathrm{K}$ for a temperature difference.)

Gas constant, $R = 8.31~\mathrm{J~mol}^{-1}~\mathrm{K}^{-1}$

Step 4: Calculate the work

$W = 6 \times 8.31 \times 20$

$W = 997.2~\mathrm{J}$

Efficiency of heat engine,

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

Coefficient of performance of refrigerator,

$ \begin{aligned} & \beta=\frac{T_2}{T_1-T_2} \\ \therefore \quad & \frac{\eta}{\beta}=\frac{\frac{T_1-T_2}{T_1}}{\frac{T_2}{T_1-T_2}}=\frac{\left(T_1-T_2\right)^2}{T_1 T_2} \end{aligned} $

We are told:

$ U = 2250 R $

for

$ n = 3 \, \text{moles} $ and $ T = 27^{\circ}\mathrm{C} = 300\,\mathrm{K} $.

We are asked to find the number of degrees of freedom $ f $ of the gas.

Formula

For an ideal gas, the internal energy is given by:

$ U = \frac{f}{2} n R T $

Substitute known values

$ 2250 R = \frac{f}{2} (3) R (300) $

Cancel $ R $ from both sides:

$ 2250 = \frac{f}{2} \times 3 \times 300 $

Simplify

$ 2250 = \frac{f}{2} \times 900 $

$ 2250 = 450 f $

$ f = \frac{2250}{450} = 5 $

✅ Answer: $ f = 5 $

That corresponds to Option B.

Interpretation: The gas has 5 degrees of freedom — likely a diatomic gas (3 translational + 2 rotational).

According to the principle of calorimetry,

$ \begin{aligned} & \quad Q_{\text {gained }}=Q_{\text {lost }} \\ & \Rightarrow m_{\text {water }} \times C_{\text {water }} \times \Delta T=m_{\text {steam }} \times L_{\text {steam }} +m_{\text {steam }} \times C_{\text {water }} \times \Delta T \\ & \Rightarrow 114 \times 1 \times(70-30)=m_{\text {steam }} \times 540 +m_{\text {steam }} \times 1 \times 30 \\ & \Rightarrow m_{\text {steam }}=8 \mathrm{~g} \\ & \therefore \text { Total mass of water }=m_{\text {water }}+m_{\text {steam }} \\ & =114+8=122 \mathrm{~g} \end{aligned} $

If $W_1$ and $W_2$ represents work done during isothermal compression and isothermal expansion.

Then, $\quad W_2=W_1+25 \%$ of $W_1$

$ \begin{aligned} & =W_1+\frac{25}{100} W_1=\frac{5 W_1}{4} \\ \therefore \quad Q_2 & =\frac{5 Q_1}{4} \end{aligned} $

$ \begin{aligned}Thus, \eta & =1-\frac{W_1}{W_2}=1-\frac{Q_1}{Q_2}=1-\frac{Q_1}{\frac{5 Q_1}{4}} \\ & =1-\frac{4}{5}=\frac{1}{5}=0.20 \text { or } 20 \% \end{aligned} $

Formula for Work in Isothermal Process

The work done when a gas changes its volume at constant temperature (isothermal process) is given by:

$ W = nRT \ln \left(\frac{V_2}{V_1}\right) $

First Case: From $ V $ to $ 2V $

We have $ n = 2 $, $ V_1 = V $, and $ V_2 = 2V $.

So, the work done is:

$ W = 2RT \ln\left(\frac{2V}{V}\right) = 2RT \ln(2) \quad ...(i) $

Second Case: From $ 2V $ to $ 4V $

Now, $ n = 2 $, $ V_1 = 2V $, and $ V_2 = 4V $.

So, the work done here is:

$ W' = 2RT \ln\left(\frac{4V}{2V}\right) $

$ = 2RT \ln(2) = W $

The work done in increasing the volume from $ 2V $ to $ 4V $ is the same as the work done from $ V $ to $ 2V $, both are $ W $.

Step 1: Write the adiabatic equation

An adiabatic process for an ideal gas can be written as:

$ p V^\gamma = \text{constant} $

Step 2: Take logarithms on both sides

If we take the logarithm of both sides, we get:

$ \log(p) + \gamma \log(V) = \text{constant} $

This can be rearranged as:

$ \log(p) = -\gamma \log(V) + \text{constant} $

Step 3: Compare to the standard line equation

This equation looks like the equation of a straight line:

$ y = m x + c $

Here, $y = \log(p)$, $x = \log(V)$, and the slope $m = -\gamma$.

Step 4: Find the slope from the graph

The slope of the line is:

$ m = \frac{\Delta \log p}{\Delta \log V} $

From the graph, pick two points and subtract their values:

$ = \frac{2.20 - 2.48}{1.4 - 1.2} = \frac{-0.28}{0.2} = -1.4 $

Step 5: Interpret the value of $\gamma$

The slope $m = -\gamma = -1.4$, so:

$ \gamma = 1.4 $

Step 6: Final answer

The ratio of the specific heat capacities of the gas is $1.4$.

$U=\frac{f}{2} n R T$

For one mole gas, $n=1$

For rigid diatomic gas, $f=5$

$ \therefore \quad U=\frac{5}{2} R T $

According to Wien's displacement law.

$ \begin{aligned} & \lambda_m \cdot T=b \\ \Rightarrow & \lambda_m \propto \frac{1}{T} \Rightarrow \frac{\left(\lambda_m\right)_A}{\left(\lambda_m\right)_B}=\frac{T_B}{T_A} \\ \Rightarrow & \frac{0.5 \times 10^{-6}}{0.1 \times 10^{-3}}=\frac{T_B}{T_A} \\ \Rightarrow & \frac{5 \times 10^{-3}}{1}=\frac{T_B}{T_A} \\ \Rightarrow & \frac{T_A}{T_B}=\frac{1}{5 \times 10^{-3}}=\frac{10^3}{5}=\frac{1000}{5}=200 \end{aligned} $