Heat Engine and Carnot Cycle Questions in English

(C) The efficiency $\eta$ of a Carnot heat engine is given by the formula $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source (input) temperature and $T_2$ is the sink (exhaust) temperature in Kelvin.
For the engine to be $100 \%$ efficient,$\eta$ must be equal to $1$.
Substituting $\eta = 1$ into the equation: $1 = 1 - \frac{T_2}{T_1}$.
This implies $\frac{T_2}{T_1} = 0$,which means $T_2 = 0 \, K$.
Therefore,a heat engine can be $100 \%$ efficient only if its exhaust temperature is absolute zero $(0 \, K)$.

(B) According to Carnot's theorem,the efficiency of all reversible engines operating between the same two temperatures is the same.
Furthermore,the efficiency of any irreversible engine operating between the same two temperatures is always less than the efficiency of a reversible engine.
Therefore,the efficiency of the reversible engine is greater than that of the irreversible engine.

(A) In a standard Carnot cycle represented on a $P-V$ diagram:
$1$. The process $A B$ represents an isothermal expansion at a higher temperature $T_1$.
$2$. The process $B C$ represents an adiabatic expansion.
$3$. The process $C D$ represents an isothermal compression at a lower temperature $T_2$.
$4$. The process $D A$ represents an adiabatic compression.
Therefore,the portion $A B$ represents the isothermal expansion.

(A) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Given $T_1 = 427^{\circ} C = 427 + 273 = 700\,K$ and $T_2 = 27^{\circ} C = 27 + 273 = 300\,K$.
$\eta = 1 - \frac{300}{700} = 1 - \frac{3}{7} = \frac{4}{7}$.
Efficiency is also defined as $\eta = \frac{W}{Q_1}$,where $W$ is the work done per second $(1.0\,kW = 1000\,J/s)$ and $Q_1$ is the heat absorbed per second.
$Q_1 = \frac{W}{\eta} = \frac{1000\,J/s}{4/7} = 1000 \times \frac{7}{4} = 1750\,J/s$.
To convert $J/s$ (Watts) to $kcal/s$,we use the conversion $1\,kcal = 4184\,J$ (or approximately $4200\,J$).
$Q_1 = \frac{1750}{4184} \approx 0.418\,kcal/s$. Given the options,$0.417\,kcal/s$ is the correct value.

(A) Case $1$: The engine operates between $T_H = 200^{\circ} C = 473 \ K$ and $T_L = 0^{\circ} C = 273 \ K$.
The efficiency is $\eta_1 = 1 - \frac{T_L}{T_H} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.4228$.
Case $2$: The engine operates between $T_H = 0^{\circ} C = 273 \ K$ and $T_L = -200^{\circ} C = 73 \ K$.
The efficiency is $\eta_2 = 1 - \frac{T_L}{T_H} = 1 - \frac{73}{273} = \frac{200}{273} \approx 0.7326$.
The ratio of efficiencies is $\frac{\eta_1}{\eta_2} = \frac{200/473}{200/273} = \frac{273}{473} \approx 0.577$.

(B) Initially,the efficiency $\eta_1 = 50\,\% = 0.5$. The source temperature $T_1 = 600\,K$.
Using the Carnot efficiency formula $\eta = 1 - \frac{T_2}{T_1}$:
$0.5 = 1 - \frac{T_2}{600}$
$\frac{T_2}{600} = 0.5$
$T_2 = 300\,K$ (This is the temperature of the sink).
Now,the efficiency is increased to $\eta_2 = 70\,\% = 0.7$,while the sink temperature $T_2$ remains $300\,K$. Let the new source temperature be $T_1'$.
Using the formula again:
$0.7 = 1 - \frac{300}{T_1'}$
$\frac{300}{T_1'} = 1 - 0.7 = 0.3$
$T_1' = \frac{300}{0.3} = 1000\,K$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the temperature of the cold reservoir and $T_1$ is the temperature of the hot reservoir.
As $T_2$ decreases,the ratio $\frac{T_2}{T_1}$ decreases,which leads to an increase in efficiency $\eta$.
The lowest possible temperature for the cold reservoir is absolute zero,which is $-273.15^{\circ} C$ (often approximated as $-273^{\circ} C$).
At $T_2 = 0 \ K$,the efficiency $\eta$ becomes $1 - 0 = 1$ or $100\%$,which is the maximum possible efficiency.
Therefore,Assertion $A$ is true.
Reason $R$ correctly states the formula for efficiency and explains that it depends on both temperatures,which justifies why lowering $T_2$ increases the efficiency.
Thus,both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.
Given $T_H = 99^{\circ} C = 99 + 273 = 372 K$.
For the first case,$\eta_1 = \frac{1}{3} = 1 - \frac{T_C}{372}$.
$\frac{T_C}{372} = 1 - \frac{1}{3} = \frac{2}{3}$.
$T_C = \frac{2}{3} \times 372 = 248 K$.
For the second case,the cold reservoir temperature is increased by $x$,so $T_C' = T_C + x = 248 + x$.
The new efficiency is $\eta_2 = \frac{1}{6} = 1 - \frac{248 + x}{372}$.
$\frac{248 + x}{372} = 1 - \frac{1}{6} = \frac{5}{6}$.
$248 + x = \frac{5}{6} \times 372 = 5 \times 62 = 310$.
$x = 310 - 248 = 62 K$.

(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.
Given temperatures are $T_1 = 127^{\circ}C = 127 + 273 = 400\,K$ and $T_2 = 27^{\circ}C = 27 + 273 = 300\,K$.
Efficiency $\eta = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$.
Also,efficiency is defined as $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat absorbed from the source.
Given $W = 2\,kJ$,we have $0.25 = \frac{2\,kJ}{Q_1}$.
Therefore,$Q_1 = \frac{2}{0.25} = 8\,kJ$.

(A) The temperatures are $T_1 = 373 \text{ K}$ (boiling point) and $T_2 = 273 \text{ K}$ (freezing point).
The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{373} \approx 0.268$ or $26.8 \%$.
Any real engine operating between these two temperatures must have an efficiency less than the Carnot efficiency $(\eta < 26.8 \%)$.
Therefore,the efficiency is less than $26.8 \%$,which implies it is less than $27 \%$.
Thus,statements $2$ and $4$ are correct.

(B) The efficiency of a Carnot engine is given by the formula:
$\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$
Given efficiency $\eta = 50\% = 0.5$.
The temperature of the source $T_{\text{source}} = 327^{\circ}C = 327 + 273 = 600\,K$.
Substituting the values into the formula:
$0.5 = 1 - \frac{T_{\text{sink}}}{600}$
$\frac{T_{\text{sink}}}{600} = 1 - 0.5 = 0.5$
$T_{\text{sink}} = 0.5 \times 600 = 300\,K$.
To convert the temperature back to Celsius:
$T_{\text{sink}}(^{\circ}C) = 300 - 273 = 27^{\circ}C$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.
For engine $E$: $\eta_{12} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.423$.
For engine $E_1$: $\eta_1 = 1 - \frac{373}{473} = \frac{100}{473} \approx 0.211$.
For engine $E_2$: $\eta_2 = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268$.
Calculating the sum: $\eta_1 + \eta_2 = 0.211 + 0.268 = 0.479$.
Comparing the values: $0.423 < 0.479$,therefore $\eta_{12} < \eta_1 + \eta_2$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Given $\eta = \frac{1}{5}$,we have $\frac{1}{5} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{4}{5}$ or $T_2 = 0.8 T_1$.
When $T_2$ is lowered by $45 \ K$,the new temperature is $T_2' = T_2 - 45$.
The new efficiency is $\eta' = \frac{1}{2}$,so $\frac{1}{2} = 1 - \frac{T_2 - 45}{T_1}$.
This simplifies to $\frac{T_2 - 45}{T_1} = \frac{1}{2}$,or $T_2 - 45 = 0.5 T_1$.
Substitute $T_2 = 0.8 T_1$ into the equation: $0.8 T_1 - 45 = 0.5 T_1$.
$0.3 T_1 = 45$,which gives $T_1 = \frac{45}{0.3} = 150 \ K$.
Now,find $T_2$: $T_2 = 0.8 \times 150 = 120 \ K$.
Thus,the temperatures are $150 \ K$ and $120 \ K$.

(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Case $1$: $\eta_1 = \frac{1}{6} = 1 - \frac{T_2}{T_1} \implies \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \implies T_2 = \frac{5}{6} T_1$.
Case $2$: $\eta_2 = \frac{1}{3} = 1 - \frac{T_2 - 57}{T_1} \implies \frac{T_2 - 57}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \implies T_2 - 57 = \frac{2}{3} T_1$.
Substitute $T_2 = \frac{5}{6} T_1$ into the second equation:
$\frac{5}{6} T_1 - 57 = \frac{2}{3} T_1$
$\frac{5}{6} T_1 - \frac{4}{6} T_1 = 57$
$\frac{1}{6} T_1 = 57$
$T_1 = 57 \times 6 = 342 \ K$.
Thus,the temperature of the source is $342 \ K$.

(B) The efficiency $(\eta)$ of a heat engine is defined as the ratio of the output power $(P_{\text{out}})$ to the input power $(P_{\text{in}})$.
$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$
Given,$\eta = 20 \% = 0.20$ and $P_{\text{in}} = 2 \text{ kW} = 2000 \text{ W}$.
Substituting the values into the formula:
$0.20 = \frac{P_{\text{out}}}{2000 \text{ W}}$
$P_{\text{out}} = 0.20 \times 2000 \text{ W} = 400 \text{ W}$.
Therefore,the output power of the engine is $400 \text{ W}$.

(C) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Thus,$T_H V_1^{\gamma-1} = T_C V_2^{\gamma-1}$,where $T_H$ is the source temperature and $T_C$ is the sink temperature.
Given that the gas is diatomic,the adiabatic index $\gamma = 1.4$.
The volume changes from $V_1 = V$ to $V_2 = 32V$.
Substituting these values:
$\frac{T_C}{T_H} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} = \left(\frac{V}{32V}\right)^{1.4-1} = \left(\frac{1}{32}\right)^{0.4}$.
Since $32 = 2^5$,we have $\left(\frac{1}{2^5}\right)^{0.4} = \left(\frac{1}{2^5}\right)^{2/5} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25$.
The efficiency of the Carnot engine is $\eta = 1 - \frac{T_C}{T_H}$.
$\eta = 1 - 0.25 = 0.75$.

(C) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source temperature.
Given $\eta_1 = 40 \% = 0.4$ and $T_{H1} = 600 \ K$:
$0.4 = 1 - \frac{T_L}{600}$
$\frac{T_L}{600} = 1 - 0.4 = 0.6$
$T_L = 0.6 \times 600 = 360 \ K$.
Now,for the desired efficiency $\eta_2 = 60 \% = 0.6$ with the same sink temperature $T_L = 360 \ K$:
$0.6 = 1 - \frac{360}{T_{H2}}$
$\frac{360}{T_{H2}} = 1 - 0.6 = 0.4$
$T_{H2} = \frac{360}{0.4} = 900 \ K$.

(B) The efficiency $(\eta)$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_C}{T_H}$.
Substituting the given values $T_H = 600 \ K$ and $T_C = 300 \ K$:
$\eta = 1 - \frac{300}{600} = 1 - 0.5 = 0.5$.
Efficiency is also defined as the ratio of work done $(W)$ to the heat absorbed from the source $(Q)$:
$\eta = \frac{W}{Q}$.
Given $W = 1.5 \ kJ$,we can solve for $Q$:
$0.5 = \frac{1.5}{Q}$.
$Q = \frac{1.5}{0.5} = 3.0 \ kJ$.
Therefore,the heat transferred to the engine is $3.0 \ kJ$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Given $\eta_1 = \frac{1}{6}$,we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{5}{6}$,or $T_2 = \frac{5}{6} T_1$ (Equation $i$).
When $T_2$ is lowered by $62 \ K$,the new temperature is $(T_2 - 62) \ K$. The new efficiency is $\eta_2 = \frac{1}{3}$.
Thus,$\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$.
Rearranging gives $\frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3}$.
Substituting $T_2 = \frac{5}{6} T_1$ into this equation:
$\frac{\frac{5}{6} T_1 - 62}{T_1} = \frac{2}{3}$.
$\frac{5}{6} - \frac{62}{T_1} = \frac{2}{3}$.
$\frac{62}{T_1} = \frac{5}{6} - \frac{2}{3} = \frac{5-4}{6} = \frac{1}{6}$.
Therefore,$T_1 = 62 \times 6 = 372 \ K$.
Now,$T_2 = \frac{5}{6} \times 372 = 310 \ K$.

(A) The Carnot cycle consists of four reversible processes:
$1$. Isothermal expansion: The gas expands at a constant high temperature $T_H$ by absorbing heat $Q_H$ from the source.
$2$. Adiabatic expansion: The gas expands further without any heat exchange,and its temperature drops to $T_L$.
$3$. Isothermal compression: The gas is compressed at a constant low temperature $T_L$ while rejecting heat $Q_L$ to the sink.
$4$. Adiabatic compression: The gas is compressed back to its initial state without heat exchange,and its temperature rises back to $T_H$.
Therefore,the first operation is isothermal expansion.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.
Given initial efficiency $\eta_1 = 50 \% = 0.5$ and source temperature $T_{H1} = 600 \ K$.
$0.5 = 1 - \frac{T_C}{600} \implies \frac{T_C}{600} = 0.5 \implies T_C = 300 \ K$.
Now,the efficiency is increased to $\eta_2 = 70 \% = 0.7$ while keeping the sink temperature $T_C = 300 \ K$ constant.
$0.7 = 1 - \frac{300}{T_{H2}}$.
$\frac{300}{T_{H2}} = 1 - 0.7 = 0.3$.
$T_{H2} = \frac{300}{0.3} = 1000 \ K$.
Thus,the new temperature of the source is $1000 \ K$.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.
For case $(i)$,the sink temperature $T_C = 100 \ K$ and the source temperature $T_H = 600 \ K$.
Therefore,$\eta_1 = 1 - \frac{100}{600} = 1 - \frac{1}{6} = \frac{5}{6}$.
For case $(ii)$,the sink temperature $T_C = T \ K$ and the source temperature $T_H = 960 \ K$.
Therefore,$\eta_2 = 1 - \frac{T}{960}$.
Since the efficiency is the same in both cases,$\eta_1 = \eta_2$.
$\frac{5}{6} = 1 - \frac{T}{960}$.
$\frac{T}{960} = 1 - \frac{5}{6} = \frac{1}{6}$.
$T = \frac{960}{6} = 160 \ K$.

(D) The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$
Here,the temperature of the source $T_1 = 227^{\circ}C = 227 + 273 = 500 \ K$.
The temperature of the sink $T_2 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Substituting these values: $\eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$ or $\frac{2}{5}$.
We know that efficiency $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat supplied.
Given $Q_1 = 50 \ kJ$,we have $W = \eta \times Q_1$.
$W = 0.4 \times 50 \ kJ = 20 \ kJ$.

(D) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
For option $A$ $(T_1 = 60 \ K, T_2 = 40 \ K)$: $\eta = 1 - \frac{40}{60} = 1 - 0.667 = 0.333$.
For option $B$ $(T_1 = 40 \ K, T_2 = 20 \ K)$: $\eta = 1 - \frac{20}{40} = 1 - 0.5 = 0.5$.
For option $C$ $(T_1 = 80 \ K, T_2 = 60 \ K)$: $\eta = 1 - \frac{60}{80} = 1 - 0.75 = 0.25$.
For option $D$ $(T_1 = 100 \ K, T_2 = 80 \ K)$: $\eta = 1 - \frac{80}{100} = 1 - 0.8 = 0.2$.
Comparing the values,the efficiency is the least for the combination $T_1 = 100 \ K$ and $T_2 = 80 \ K$.

(D) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = \left(1 - \frac{T_{2}}{T_{1}}\right) \times 100 \%$.
Given: $T_{1} = 500 \ K$ (source temperature) and $T_{2} = 300 \ K$ (sink temperature).
Substituting the values into the formula:
$\eta = \left(1 - \frac{300}{500}\right) \times 100 \%$.
$\eta = \left(1 - 0.6\right) \times 100 \%$.
$\eta = 0.4 \times 100 \% = 40 \%$.
Thus,the efficiency of the Carnot engine is $40 \%$.

(D) Given,efficiency,$\eta = 70 \% = 0.7$.
Sink temperature,$T_2 = 27^{\circ}C = 273 + 27 = 300 \ K$.
The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.
Substituting the given values: $0.7 = 1 - \frac{300}{T_1}$.
Rearranging the terms: $\frac{300}{T_1} = 1 - 0.7 = 0.3$.
Solving for source temperature $T_1$: $T_1 = \frac{300}{0.3} = 1000 \ K$.
Converting the temperature back to Celsius: $T_1 = 1000 - 273 = 727^{\circ}C$.
Therefore,the source temperature required is $727^{\circ}C$.

(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.
From the $p-V$ diagram,the process $AB$ is isothermal expansion at $T_1$ and $CD$ is isothermal compression at $T_2$.
For an isothermal process,$pV = \text{constant}$.
At point $A$: $p_A = 1600 \ kPa$,$V_A = 2.5 \ cm^3$. So,$p_A V_A = 1600 \times 2.5 = 4000$.
At point $C$: $p_C = 400 \ kPa$,$V_C = 6.25 \ cm^3$. So,$p_C V_C = 400 \times 6.25 = 2500$.
Since $pV = \mu RT$,we have $\frac{T_2}{T_1} = \frac{p_C V_C}{p_A V_A} = \frac{2500}{4000} = \frac{5}{8} = 0.625$.
The efficiency is $\eta = 1 - 0.625 = 0.375$.
Given heat supplied $Q_1 = 8000 \ J$.
Work done $W = \eta \times Q_1 = 0.375 \times 8000 = 3000 \ J$.

(B) The efficiency $(\eta)$ of a Carnot engine is given by the formula:
$\eta = 1 - \frac{T_{L}}{T_{H}}$
where $T_{L}$ is the temperature of the cold reservoir and $T_{H}$ is the temperature of the hot reservoir.
For a fixed cold reservoir temperature $T_{L}$,as the hot reservoir temperature $T_{H}$ increases,the ratio $\frac{T_{L}}{T_{H}}$ decreases.
Consequently,the efficiency $\eta = 1 - \frac{T_{L}}{T_{H}}$ increases as $T_{H}$ increases.
When $T_{H} = T_{L}$,the efficiency $\eta = 1 - \frac{T_{L}}{T_{L}} = 0$.
As $T_{H} \to \infty$,the efficiency $\eta \to 1$.
The graph of $\eta$ versus $T_{H}$ starts from $0$ at $T_{H} = T_{L}$ and increases with a decreasing slope (concave down) towards the value $1$.
This behavior is correctly represented by the graph in option $(b)$.

(B) Given: Heat taken from source,$Q_{1} = 300$ calories; Temperature of source,$T_{1} = 500 \,K$; Heat rejected to the sink,$Q_{2} = 150$ calories.
We need to find the temperature of the sink,$T_{2}$.
For a Carnot engine,the efficiency $\eta$ is given by the ratio of heat exchange as $\eta = \frac{Q_{1} - Q_{2}}{Q_{1}}$ and also by the temperatures as $\eta = \frac{T_{1} - T_{2}}{T_{1}}$.
Equating the two expressions for efficiency:
$\frac{Q_{1} - Q_{2}}{Q_{1}} = \frac{T_{1} - T_{2}}{T_{1}}$
Substituting the given values:
$\frac{300 - 150}{300} = \frac{500 - T_{2}}{500}$
$\frac{150}{300} = 1 - \frac{T_{2}}{500}$
$\frac{1}{2} = 1 - \frac{T_{2}}{500}$
$\frac{T_{2}}{500} = 1 - \frac{1}{2}$
$\frac{T_{2}}{500} = \frac{1}{2}$
$T_{2} = 500 \times \frac{1}{2} = 250 \,K$.
Thus,the temperature of the sink is $250 \,K$.

(B) Given:
Temperature of source,$T_1 = 400 \ K$
Temperature of sink,$T_2 = 300 \ K$
Useful work done,$W = 800 \ J$
Efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Also,efficiency is defined as $\eta = \frac{W}{Q_1}$,where $Q_1$ is the heat supplied.
Equating the two,we get $1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.
Substituting the values: $1 - \frac{300}{400} = \frac{800}{Q_1}$.
$\frac{100}{400} = \frac{800}{Q_1}$.
$\frac{1}{4} = \frac{800}{Q_1}$.
$Q_1 = 800 \times 4 = 3200 \ J$.
Thus,the heat energy supplied to the engine is $3200 \ J$.

(D) The efficiency of a Carnot heat engine is given by the formula $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.
For option $A$: $\eta_A = 1 - \frac{400}{600} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.333$.
For option $B$: $\eta_B = 1 - \frac{200}{400} = 1 - \frac{1}{2} = \frac{1}{2} = 0.500$.
For option $C$: $\eta_C = 1 - \frac{300}{500} = 1 - \frac{3}{5} = \frac{2}{5} = 0.400$.
For option $D$: $\eta_D = 1 - \frac{100}{300} = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.667$.
Comparing the values,$\eta_D$ is the maximum efficiency.
Therefore,the correct combination is $T_1 = 300 \ K$ and $T_2 = 100 \ K$.

(A) The efficiency $\eta$ of a Carnot heat engine is given by the formula: $\eta = 1 - \frac{T_{2}}{T_{1}}$.
Given that $\eta = 0.5$, we have $0.5 = 1 - \frac{T_{2}}{T_{1}}$, which implies $\frac{T_{2}}{T_{1}} = 0.5$, or $T_{1} = 2T_{2}$.
For another Carnot engine to have the same efficiency of $0.5$, the ratio of the sink temperature to the source temperature must remain the same, i.e., $\frac{T_{2}'}{T_{1}'} = 0.5$.
If we multiply both the source and sink temperatures by the same factor $k$, the new efficiency is $\eta' = 1 - \frac{k T_{2}}{k T_{1}} = 1 - \frac{T_{2}}{T_{1}} = 0.5$.
Therefore, if the source temperature is $2T_{1}$ and the sink temperature is $2T_{2}$, the efficiency remains $0.5$.

(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink in Kelvin.
Given: $T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$ and $T_2 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Substituting the values: $\eta = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$ or $\frac{1}{4}$.
Efficiency is also defined as $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat supplied.
Given $Q_1 = 40 \ kJ$,we have $\frac{1}{4} = \frac{W}{40 \ kJ}$.
Therefore,$W = \frac{40}{4} \ kJ = 10 \ kJ$.

(C) Step $1$: Find the temperature where Celsius and Fahrenheit scales coincide.
Let the temperature be $T_1$. The relation is $\frac{C}{5} = \frac{F-32}{9}$. Setting $C = F = T_1$:
$\frac{T_1}{5} = \frac{T_1-32}{9} \Rightarrow 9T_1 = 5T_1 - 160 \Rightarrow 4T_1 = -160 \Rightarrow T_1 = -40^{\circ}C$.
In Kelvin,$T_1 = -40 + 273.15 = 233.15 \ K$.
Step $2$: Find the temperature where Kelvin and Fahrenheit scales coincide.
Let the temperature be $T_2$. The relation is $\frac{K-273.15}{5} = \frac{F-32}{9}$. Setting $K = F = T_2$:
$\frac{T_2-273.15}{5} = \frac{T_2-32}{9} \Rightarrow 9T_2 - 2458.35 = 5T_2 - 160 \Rightarrow 4T_2 = 2298.35 \Rightarrow T_2 \approx 574.59 \ K$.
Step $3$: Calculate the efficiency $\eta$ of the Carnot engine.
$\eta = 1 - \frac{T_{cold}}{T_{hot}} = 1 - \frac{233.15}{574.59} \approx 1 - 0.405 = 0.595 \approx 60 \%$.

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature.
For the first case: $\eta_1 = 0.60$, $T_1 = 600 \text{ K}$.
$0.60 = 1 - \frac{T_2}{600} \implies \frac{T_2}{600} = 0.40 \implies T_2 = 240 \text{ K}$.
For the second case: $\eta_2 = 0.80$, $T_2 = 240 \text{ K}$.
$0.80 = 1 - \frac{240}{T_1'} \implies \frac{240}{T_1'} = 0.20 \implies T_1' = \frac{240}{0.20} = 1200 \text{ K}$.
Thus, the new source temperature required is $1200 \text{ K}$.

(A) For a Carnot engine,the ratio of heat exchanged is equal to the ratio of the absolute temperatures of the source and the sink: $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$.
Given:
Heat absorbed from source,$Q_1 = 600 \ J$.
Heat rejected to sink,$Q_2 = 400 \ J$.
Temperature of source,$T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$.
Using the formula: $\frac{600}{400} = \frac{400}{T_2}$.
Simplifying the ratio: $1.5 = \frac{400}{T_2}$.
Solving for $T_2$: $T_2 = \frac{400}{1.5} = 266.67 \ K \approx 266.7 \ K$.

(B) For a Carnot engine, the efficiency $\eta$ is given by $\eta = 1 - \frac{T_{low}}{T_{high}}$.
Also, the work done $W$ is given by $W = Q_{in} \cdot \eta$, where $Q_{in}$ is the heat absorbed from the source.
For engine $A$: $W_A = Q_A \left(1 - \frac{T}{600}\right)$.
For engine $B$: $W_B = Q_B \left(1 - \frac{400}{T}\right)$.
Since the engines are in series, the heat rejected by engine $A$ is the heat absorbed by engine $B$, so $Q_B = Q_A \left(\frac{T}{600}\right)$.
Given $W_A = W_B$, we have $Q_A \left(1 - \frac{T}{600}\right) = Q_A \left(\frac{T}{600}\right) \left(1 - \frac{400}{T}\right)$.
Simplifying the equation: $1 - \frac{T}{600} = \frac{T}{600} - \frac{400}{600}$.
$1 + \frac{400}{600} = \frac{2T}{600}$.
$1 + \frac{2}{3} = \frac{T}{300} \implies \frac{5}{3} = \frac{T}{300}$.
$T = \frac{5 \times 300}{3} = 500 \ K$.

(C) In a Carnot engine,the work done during isothermal expansion $(W_{exp})$ is given by $W_{exp} = nRT_H \ln(V_2/V_1)$ and the work done during isothermal compression $(W_{comp})$ is given by $W_{comp} = nRT_L \ln(V_2/V_1)$.
Given that $W_{exp} = W_{comp} + 0.25 W_{comp} = 1.25 W_{comp}$.
Substituting the expressions,we get $nRT_H \ln(V_2/V_1) = 1.25 nRT_L \ln(V_2/V_1)$.
This simplifies to $T_H = 1.25 T_L$,or $T_L/T_H = 1/1.25 = 0.8$.
The efficiency of a Carnot engine is given by $\eta = 1 - (T_L/T_H)$.
Substituting the value,$\eta = 1 - 0.8 = 0.2$.
Thus,the efficiency is $20 \%$.

(C) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.
Given that the source temperature $T_1$ is $25 \%$ more than the sink temperature $T_2$,we have: $T_1 = T_2 + 0.25 T_2 = 1.25 T_2$.
Substituting $T_1$ into the efficiency formula: $\eta = 1 - \frac{T_2}{1.25 T_2} = 1 - \frac{1}{1.25}$.
Since $1.25 = \frac{5}{4}$,we get: $\eta = 1 - \frac{4}{5} = \frac{1}{5}$.
Converting this to percentage: $\eta = \frac{1}{5} \times 100 \% = 20 \%$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Let $\Delta T = T_1 - T_2$ be the temperature difference,which is constant for both engines.
Thus,$\eta = \frac{\Delta T}{T_1}$.
Given the ratio of efficiencies $\frac{\eta_A}{\eta_B} = 1.25 = \frac{5}{4}$.
Since $\eta_A = \frac{\Delta T}{T_{1A}}$ and $\eta_B = \frac{\Delta T}{T_{1B}}$,we have $\frac{\eta_A}{\eta_B} = \frac{T_{1B}}{T_{1A}} = \frac{5}{4}$.
Therefore,the ratio of the absolute temperatures of the sources $T_{1A} : T_{1B} = 4 : 5$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Given $\eta_1 = \frac{1}{3}$,so $1 - \frac{T_2}{T_1} = \frac{1}{3} \implies \frac{T_2}{T_1} = \frac{2}{3} \implies T_1 = 1.5 T_2$.
When the source temperature is decreased by $50 \ K$ and the sink temperature is increased by $25 \ K$,the new efficiency is $\eta_2 = \frac{3}{16}$.
Thus,$1 - \frac{T_2 + 25}{T_1 - 50} = \frac{3}{16} \implies \frac{T_2 + 25}{T_1 - 50} = \frac{13}{16}$.
Substituting $T_1 = 1.5 T_2$ into the equation:
$\frac{T_2 + 25}{1.5 T_2 - 50} = \frac{13}{16}$.
Cross-multiplying: $16(T_2 + 25) = 13(1.5 T_2 - 50)$.
$16 T_2 + 400 = 19.5 T_2 - 650$.
$1050 = 3.5 T_2$.
$T_2 = \frac{1050}{3.5} = 300 \ K$.

(D) Let the absolute temperature of the sink be $T_2 = T$.
The absolute temperature of the source $T_1$ is $25 \%$ more than the sink temperature,so $T_1 = T + 0.25T = 1.25T$.
The efficiency $\eta$ of a Carnot engine is given by the formula $\eta = 1 - \frac{T_2}{T_1}$.
Substituting the values,we get $\eta = 1 - \frac{T}{1.25T} = 1 - \frac{1}{1.25} = 1 - \frac{100}{125} = 1 - \frac{4}{5} = \frac{1}{5}$.
To express this as a percentage,$\eta = \frac{1}{5} \times 100 \% = 20 \%$.

(C) For a Carnot cycle,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.
Given temperatures in Kelvin: $T_1 = 127 + 273 = 400 \text{ K}$ and $T_2 = 27 + 273 = 300 \text{ K}$.
Given heat absorbed $Q_1 = 5 \times 10^4 \text{ cal}$.
Substituting the values into the efficiency formula:
$\eta = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$.
Since $\eta = \frac{W}{Q_1}$,we have $W = \eta \times Q_1$.
$W = 0.25 \times 5 \times 10^4 \text{ cal} = 1.25 \times 10^4 \text{ cal}$.
Thus,the amount of heat converted to work is $1.25 \times 10^4 \text{ cal}$.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{sink}}{T_{source}}$.
For the first case,$T_1 = 800 \ K$ and $T_2 = 500 \ K$.
$\eta_1 = 1 - \frac{500}{800} = 1 - \frac{5}{8} = \frac{3}{8}$.
For the second case,$T_{source} = x \ K$ and $T_{sink} = 600 \ K$.
$\eta_2 = 1 - \frac{600}{x}$.
Since the efficiencies are equal,$\eta_1 = \eta_2$.
$1 - \frac{600}{x} = \frac{3}{8}$.
$\frac{600}{x} = 1 - \frac{3}{8} = \frac{5}{8}$.
$x = \frac{600 \times 8}{5} = 120 \times 8 = 960 \ K$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
For the first case: $\frac{1}{6} = 1 - \frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1} = \frac{5}{6} \Rightarrow T_2 = \frac{5}{6} T_1$ ....$(i)$
For the second case,the sink temperature is reduced by $65 \ K$,so the new sink temperature is $(T_2 - 65)$.
$\frac{1}{3} = 1 - \frac{T_2 - 65}{T_1} \Rightarrow \frac{T_2 - 65}{T_1} = \frac{2}{3}$ ....(ii)
Substituting $T_2 = \frac{5}{6} T_1$ from equation $(i)$ into equation (ii):
$\frac{\frac{5}{6} T_1 - 65}{T_1} = \frac{2}{3} \Rightarrow \frac{5}{6} - \frac{65}{T_1} = \frac{2}{3}$
$\frac{65}{T_1} = \frac{5}{6} - \frac{2}{3} = \frac{5-4}{6} = \frac{1}{6}$
$T_1 = 65 \times 6 = 390 \ K$
Now,calculating the initial sink temperature $T_2$:
$T_2 = \frac{5}{6} \times 390 = 325 \ K$
The final sink temperature is $T_2 - 65 = 325 - 65 = 260 \ K$.
Thus,the initial and final temperatures of the sink are $325 \ K$ and $260 \ K$ respectively.

(C) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Given $\eta_1 = 25\% = 0.25$ and $T_2 = 300 \ K$:
$0.25 = 1 - \frac{300}{T_1} \implies \frac{300}{T_1} = 0.75 \implies T_1 = \frac{300}{0.75} = 400 \ K$.
Now,we want the efficiency to become $\eta_2 = 50\% = 0.5$ by increasing the source temperature by $x$:
$0.5 = 1 - \frac{300}{400 + x} \implies \frac{300}{400 + x} = 0.5 \implies 400 + x = \frac{300}{0.5} = 600 \ K$.
$x = 600 - 400 = 200 \ K$.
Thus,the required increase in the source temperature is $200 \ K$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{sink}}{T_{source}}$,where temperatures must be in Kelvin $(K = ^{\circ}C + 273)$.
Case $1$: $T_{source} = 200^{\circ}C = 473 \ K$ and $T_{sink} = 0^{\circ}C = 273 \ K$.
$\eta_1 = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$.
Case $2$: $T_{source} = 0^{\circ}C = 273 \ K$ and $T_{sink} = -200^{\circ}C = 73 \ K$.
$\eta_2 = 1 - \frac{73}{273} = \frac{273 - 73}{273} = \frac{200}{273}$.
Now,calculating the ratio:
$\frac{\eta_1}{\eta_2} = \frac{200}{473} \times \frac{273}{200} = \frac{273}{473} \approx 0.577 \approx 0.58$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Initially,$\eta_1 = 25 \% = 0.25$.
So,$0.25 = 1 - \frac{T_2}{T_1} \implies \frac{T_2}{T_1} = 0.75 = \frac{3}{4}$.
This gives $T_1 = \frac{4}{3} T_2$.
When the source temperature is increased by $100 \ K$,the new efficiency $\eta_2 = 40 \% = 0.4$.
So,$0.4 = 1 - \frac{T_2}{T_1 + 100}$.
This implies $\frac{T_2}{T_1 + 100} = 0.6 = \frac{3}{5}$.
Substituting $T_1 = \frac{4}{3} T_2$ into the equation:
$\frac{T_2}{\frac{4}{3} T_2 + 100} = \frac{3}{5}$.
$5 T_2 = 3 \left( \frac{4}{3} T_2 + 100 \right)$.
$5 T_2 = 4 T_2 + 300$.
$T_2 = 300 \ K$.

(C) For a Carnot engine,efficiency $\eta = 1 - \frac{T_2}{T_1}$.
Given $\eta = 25 \% = 0.25$ and $T_1 = 127^{\circ} C = (127 + 273) K = 400 K$.
Substituting these values: $0.25 = 1 - \frac{T_2}{400}$.
$\frac{T_2}{400} = 1 - 0.25 = 0.75$.
$T_2 = 0.75 \times 400 = 300 K$.
In the second case,the sink temperature is decreased by $10 \%$,so $T_2' = T_2 - 0.10 T_2 = 0.9 T_2$.
$T_2' = 0.9 \times 300 = 270 K$.
The new efficiency $\eta' = 1 - \frac{T_2'}{T_1} = 1 - \frac{270}{400}$.
$\eta' = 1 - 0.675 = 0.325 = 32.5 \%$.