Heat Engine and Carnot Cycle Questions in English

(D) heat engine is a device that converts heat energy into mechanical work through a cyclic process.
$1$. It absorbs heat from a high-temperature source $(T_H)$.
$2$. It performs some amount of work $(W)$.
$3$. It rejects the remaining heat to a low-temperature sink $(T_L)$.
$4$. Since it operates in a cyclic process, the change in internal energy $(\Delta U)$ of the system over one complete cycle is zero $(\Delta U = 0)$.
Therefore, all the mentioned features are characteristic of a heat engine.

(N/A) The efficiency $\eta$ of a heat engine is defined as the ratio of the net work done $(W)$ by the engine per cycle to the total heat absorbed $(Q_H)$ from the hot reservoir (source) per cycle.
It represents the fraction of heat energy converted into useful work.
The formula for efficiency is given by:
$\eta = \frac{W}{Q_H}$
Since $W = Q_H - Q_L$ (where $Q_L$ is the heat rejected to the cold reservoir or sink), the formula can also be expressed as:
$\eta = 1 - \frac{Q_L}{Q_H}$
Efficiency is typically expressed as a percentage by multiplying the result by $100\%$.

(A) The theoretical efficiency of a heat engine is defined by the Carnot efficiency,which represents the maximum possible efficiency for any heat engine operating between two temperatures,$T_1$ (source) and $T_2$ (sink).
The formula for Carnot efficiency is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the absolute temperature of the hot reservoir and $T_2$ is the absolute temperature of the cold reservoir.
This efficiency is independent of the working substance and depends only on the temperatures of the source and the sink.

(D) The efficiency of a heat engine is defined as the ratio of work done to the heat absorbed from the source. For an ideal gas undergoing a cycle,the efficiency depends on the temperatures of the source $(T_H)$ and the sink $(T_L)$. According to the Carnot theorem,the maximum efficiency of any heat engine operating between two temperatures is given by $\eta = 1 - \frac{T_L}{T_H}$. Therefore,the efficiency is not a fixed constant for an ideal gas but depends on the temperatures of the reservoirs involved in the thermodynamic process.

(N/A) Carnot cycle is a theoretical thermodynamic cycle proposed by Sadi Carnot. It describes the most efficient possible heat engine operating between two heat reservoirs at temperatures $T_{1}$ (source) and $T_{2}$ (sink),where $T_{1} > T_{2}$.
The cycle consists of four reversible processes:
$1$. Reversible isothermal expansion: The gas expands at a constant temperature $T_{1}$,absorbing heat $Q_{1}$ from the source.
$2$. Reversible adiabatic expansion: The gas expands further without heat exchange,and its temperature drops from $T_{1}$ to $T_{2}$.
$3$. Reversible isothermal compression: The gas is compressed at a constant temperature $T_{2}$,rejecting heat $Q_{2}$ to the sink.
$4$. Reversible adiabatic compression: The gas is compressed further without heat exchange,and its temperature rises from $T_{2}$ back to $T_{1}$.
The $P-V$ diagram for the Carnot cycle is shown in the figure.

(N/A) Carnot engine is a theoretical thermodynamic cycle proposed by Nicolas $L$éonard Sadi Carnot. It operates between two heat reservoirs at temperatures $T_H$ (source) and $T_L$ (sink).
$1$. Carnot Engine Components:
- $A$ cylinder with insulating walls and a perfectly conducting base.
- $A$ frictionless,insulating piston.
- $A$ working substance (ideal gas).
- $A$ heat source at high temperature $T_H$.
- $A$ heat sink at low temperature $T_L$.
- An insulating stand.
$2$. Carnot Cycle Steps:
- Step $I$: Isothermal Expansion: The gas is placed on the heat source. It expands isothermally at temperature $T_H$,absorbing heat $Q_H$ from the source.
- Step $II$: Adiabatic Expansion: The cylinder is placed on the insulating stand. The gas expands adiabatically until its temperature drops to $T_L$.
- Step $III$: Isothermal Compression: The cylinder is placed on the heat sink. The gas is compressed isothermally at temperature $T_L$,rejecting heat $Q_L$ to the sink.
- Step $IV$: Adiabatic Compression: The cylinder is placed back on the insulating stand. The gas is compressed adiabatically until its temperature rises back to $T_H$,completing the cycle.

(N/A) Carnot engine operates through a cycle of four reversible processes:
$(1)$ Isothermal expansion $(A \rightarrow B)$: The gas expands at a constant temperature $T_1$. The heat absorbed is $Q_1 = \mu RT_1 \ln(V_2/V_1)$. The work done is $W_1 = Q_1 = \mu RT_1 \ln(V_2/V_1)$.
$(2)$ Adiabatic expansion $(B \rightarrow C)$: The gas expands adiabatically from $(P_2, V_2, T_1)$ to $(P_3, V_3, T_2)$. The work done is $W_2 = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$.
$(3)$ Isothermal compression $(C \rightarrow D)$: The gas is compressed at a constant temperature $T_2$. The heat rejected is $Q_2 = \mu RT_2 \ln(V_3/V_4)$. The work done on the gas is $W_3 = \mu RT_2 \ln(V_3/V_4)$.
$(4)$ Adiabatic compression $(D \rightarrow A)$: The gas is compressed adiabatically from $(P_4, V_4, T_2)$ to $(P_1, V_1, T_1)$. The work done on the gas is $W_4 = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$.
Total work done by the system: $W = W_1 + W_2 - W_3 - W_4 = W_1 - W_3 = \mu RT_1 \ln(V_2/V_1) - \mu RT_2 \ln(V_3/V_4)$.
Efficiency $\eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$.

(N/A) Carnot's theorem consists of two main parts:
$(a)$ No heat engine operating between two given heat reservoirs at temperatures $T_{1}$ and $T_{2}$ can be more efficient than a reversible Carnot engine operating between the same two reservoirs.
$(b)$ All reversible engines operating between the same two heat reservoirs have the same efficiency,regardless of the working substance used.
To prove part $(a)$,consider an irreversible engine $I$ and a reversible Carnot engine $R$ operating between a source at temperature $T_{1}$ and a sink at temperature $T_{2}$.
Assume $I$ absorbs heat $Q_{1}$ from the source,performs work $W'$,and rejects heat $Q_{1} - W'$ to the sink. Let $R$ be coupled to $I$ such that $R$ acts as a refrigerator,taking heat $Q_{2}$ from the sink and requiring work $W$ to return heat $Q_{1}$ to the source.
If we assume $\eta_{I} > \eta_{R}$,then for the same heat $Q_{1}$ absorbed from the source,the work done by $I$ is $W' > W$. The net work output of the combined system is $W' - W$,and the net heat extracted from the cold reservoir is $(Q_{1} - W) - (Q_{1} - W') = W' - W$. This implies the system converts heat directly into work without any other effect,which violates the Kelvin-Planck statement of the second law of thermodynamics. Thus,$\eta_{I} \leq \eta_{R}$.
For part $(b)$,if we have two reversible engines $R_{1}$ and $R_{2}$,we can run one in reverse to form a combined system. If one were more efficient,it would violate the second law. Therefore,all reversible engines must have the same efficiency,given by $\eta = 1 - \frac{T_{2}}{T_{1}}$,which is independent of the working substance.

(N/A) Carnot cycle is a theoretical thermodynamic cycle proposed by Nicolas $L$éonard Sadi Carnot in $1824$. It describes the most efficient possible heat engine operating between two heat reservoirs at temperatures $T_H$ (hot) and $T_C$ (cold). The cycle consists of four reversible processes:
$1$. Reversible isothermal expansion: The gas expands at a constant high temperature $T_H$,absorbing heat $Q_H$ from the source.
$2$. Reversible adiabatic expansion: The gas continues to expand without heat exchange,and its temperature drops from $T_H$ to $T_C$.
$3$. Reversible isothermal compression: The gas is compressed at a constant low temperature $T_C$,rejecting heat $Q_C$ to the sink.
$4$. Reversible adiabatic compression: The gas is compressed without heat exchange,and its temperature rises from $T_C$ back to $T_H$,completing the cycle.

(B) Carnot engine is a theoretical model of a heat engine proposed by Sadi Carnot in $1824$.
It operates on a reversible cycle known as the Carnot cycle,which consists of two isothermal processes and two adiabatic processes.
It provides the theoretical upper limit for the efficiency of any heat engine operating between two given temperatures,$T_H$ (hot reservoir) and $T_L$ (cold reservoir).
The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where temperatures are in Kelvin.
Since it is a reversible cycle,it is the most efficient engine possible between two heat reservoirs.

(A) The Carnot engine is a theoretical thermodynamic cycle proposed by the French physicist $Nicolas$ $Léonard$ $Sadi$ $Carnot$ in $1824$.
He developed this model to determine the maximum possible efficiency that any heat engine can achieve when operating between two given thermal reservoirs at different temperatures.
Therefore,the correct answer is $Nicolas$ $Léonard$ $Sadi$ $Carnot$.

(D) An ideal engine,specifically the Carnot engine,operates on the Carnot cycle.
This cycle consists of four reversible processes:
$1$. Reversible isothermal expansion.
$2$. Reversible adiabatic expansion.
$3$. Reversible isothermal compression.
$4$. Reversible adiabatic compression.
Therefore,an ideal engine involves a combination of isothermal and adiabatic processes.

(A) 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.
From this formula,it is clear that the efficiency depends only on the temperatures of the source and the sink ($T_1$ and $T_2$).
It is independent of the nature of the working substance used in the engine and the design of the engine.

(N/A) The efficiency of a Carnot engine is given by the formula $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the absolute temperature of the cold reservoir (sink) and $T_H$ is the absolute temperature of the hot reservoir (source).
For the efficiency $\eta$ to be $100\%$ (or $1$),the term $\frac{T_L}{T_H}$ must be equal to $0$.
This implies that either $T_L = 0 \ K$ (absolute zero) or $T_H = \infty$.
According to the Third Law of Thermodynamics,it is impossible to reach absolute zero $(0 \ K)$ in a finite number of steps.
Furthermore,it is physically impossible to have a heat source at infinite temperature.
Therefore,it is impossible for any heat engine,including the ideal Carnot engine,to have an efficiency of $100\%$.

(N/A) Carnot's theorem states two main points regarding the efficiency of heat engines:
$1$. All reversible heat engines operating between the same two temperatures ($T_1$ and $T_2$) have the same efficiency,regardless of the working substance used.
$2$. No irreversible heat engine operating between two given temperatures can have an efficiency greater than that of a reversible heat engine operating between the same two temperatures.
Mathematically,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.

(N/A) An ideal heat engine,often referred to as a Carnot engine,operates on a reversible cycle. Two essential characteristics are:
$(i)$ The source of heat must have an infinite heat capacity,meaning it can supply any amount of heat without a change in its temperature.
$(ii)$ The sink must have an infinite heat capacity,meaning it can absorb any amount of heat without a change in its temperature.

(B) The efficiency 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.
For the efficiency $\eta$ to be $100 \%$ (or $1$),the term $\frac{T_2}{T_1}$ must be equal to $0$.
This condition is satisfied if the temperature of the sink $T_2$ is $0 \ K$ (absolute zero),while the source temperature $T_1$ remains finite and greater than $0 \ K$.

(B) The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.
For an isothermal process,the temperature remains constant,meaning $T_1 = T_2$.
Substituting this into the efficiency formula: $\eta = 1 - \frac{T_1}{T_1} = 1 - 1 = 0$.
Since the efficiency $\eta$ is $0$,the engine cannot convert any heat into useful work. Therefore,no useful work is performed.

(B) The efficiency of a Carnot engine is given by the formula $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the temperature of the sink and $T_H$ is the temperature of the source in Kelvin.
For water,the boiling point is $100^{\circ}C = 373.15 \ K$ and the freezing point is $0^{\circ}C = 273.15 \ K$.
Here,$T_H = 373.15 \ K$ and $T_L = 273.15 \ K$.
Substituting these values into the formula:
$\eta = 1 - \frac{273.15}{373.15} = \frac{373.15 - 273.15}{373.15} = \frac{100}{373.15} \approx 0.268$.
Converting to percentage,$\eta \approx 26.8\%$.

(A) The temperature of the source is $T_1 = 500\, K$ and the temperature of the sink is $T_2 = 300\, K$.
The work done per cycle is $W = 1\, kJ = 1000\, J$.
The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Substituting the values,$\eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$.
Since efficiency $\eta = \frac{W}{Q_1}$,the heat transferred to the engine from the source is $Q_1 = \frac{W}{\eta}$.
$Q_1 = \frac{1000\, J}{0.4} = 2500\, J$.
The heat rejected to the sink is $Q_2 = Q_1 - W = 2500\, J - 1000\, J = 1500\, J$.

(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 absolute temperature of the source and $T_{2}$ is the absolute temperature of the sink.
Therefore,the efficiency of a Carnot engine depends on both the temperature of the source and the temperature of the sink.

(B) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Rearranging this,we get $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
Given that the final volume $V_2 = 3V_1$,we have $\frac{V_1}{V_2} = \frac{1}{3}$.
Substituting the values $\gamma = 1.5$ and $\frac{V_1}{V_2} = \frac{1}{3}$:
$\frac{T_2}{T_1} = \left(\frac{1}{3}\right)^{1.5-1} = \left(\frac{1}{3}\right)^{0.5} = \frac{1}{\sqrt{3}}$.
The efficiency $\eta$ of a Carnot cycle is defined as $\eta = 1 - \frac{T_2}{T_1}$.
Substituting the ratio,we get $\eta = 1 - \frac{1}{\sqrt{3}}$.

(A) The efficiency of a Carnot cycle is given by the formula:
$\eta = 1 - \frac{T_{cold}}{T_{hot}}$
Here,the cold reservoir temperature $T_{cold} = 4^{\circ} C = 273 + 4 = 277 \ K$.
The hot reservoir temperature $T_{hot} = 15^{\circ} C = 273 + 15 = 288 \ K$.
Substituting these values into the formula:
$\eta = 1 - \frac{277}{288}$
$\eta = 1 - 0.9618$
$\eta \approx 0.0382$
Rounding to three decimal places,we get $\eta = 0.038$.

(C) For a Carnot engine,the ratio of heat exchanged is equal to the ratio of absolute temperatures:
$\frac{Q_{\text{source}}}{Q_{\text{sink}}} = \frac{T_{\text{source}}}{T_{\text{sink}}}$
Given:
$T_{\text{source}} = 127^{\circ} C = 127 + 273 = 400 \ K$
$T_{\text{sink}} = 27^{\circ} C = 27 + 273 = 300 \ K$
$Q_{\text{source}} = 500 \ J$
Substituting the values into the formula:
$\frac{500}{Q_{\text{sink}}} = \frac{400}{300}$
$\frac{500}{Q_{\text{sink}}} = \frac{4}{3}$
$Q_{\text{sink}} = \frac{500 \times 3}{4} = 125 \times 3 = 375 \ J$

(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1 = 800\, K$ (source temperature) and $T_2 = 400\, K$ (sink temperature).
$\eta = 1 - \frac{400}{800} = 1 - 0.5 = 0.5$.
Also,efficiency is defined as $\eta = \frac{W}{Q_1}$,where $W$ is the work output and $Q_1$ is the heat supplied.
Given $W = 1200\, J$,we have $0.5 = \frac{1200}{Q_1}$.
Therefore,$Q_1 = \frac{1200}{0.5} = 2400\, J$.

(A) The efficiency $\eta$ of an ideal (Carnot) heat engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_H$ is the temperature of the source and $T_L$ is the temperature of the sink in Kelvin.
Given: $\eta = 60\, \% = 0.60$ and $T_H = 127\,^{\circ} C = 127 + 273 = 400\, K$.
Substituting the values: $0.60 = 1 - \frac{T_L}{400}$.
Rearranging the equation: $\frac{T_L}{400} = 1 - 0.60 = 0.40$.
$T_L = 0.40 \times 400 = 160\, K$.
Converting back to Celsius: $T_L(^{\circ} C) = 160 - 273 = -113\,^{\circ} C$.

(B) The efficiency of a heat 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 $\eta = \frac{1}{4}$,we have $\frac{1}{4} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{3}{4}$,so $T_2 = 0.75 T_1$.
When the sink temperature is reduced by $52 \, K$,the new efficiency $\eta' = 2 \times \eta = 2 \times \frac{1}{4} = \frac{1}{2}$.
The new sink temperature is $T_2' = T_2 - 52$.
Using the efficiency formula for the new state: $\frac{1}{2} = 1 - \frac{T_2 - 52}{T_1}$.
This simplifies to $\frac{T_2 - 52}{T_1} = \frac{1}{2}$,or $T_2 - 52 = 0.5 T_1$.
Substituting $T_2 = 0.75 T_1$ into the equation: $0.75 T_1 - 52 = 0.5 T_1$.
$0.25 T_1 = 52$,which gives $T_1 = \frac{52}{0.25} = 208 \, K$.

(B) Given: Heat absorbed from the hot reservoir $Q_{\text{in}} = 300 \, J$.
Heat rejected to the cold reservoir $Q_{\text{out}} = 240 \, J$.
Temperature of the cold reservoir $T_{2} = 400 \, K$.
For a reversible heat engine (Carnot engine),the efficiency is given by $\eta = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}}$.
Substituting the values: $\eta = 1 - \frac{240}{300} = 1 - 0.8 = 0.2$.
Also,the efficiency of a Carnot engine is $\eta = 1 - \frac{T_{2}}{T_{1}}$.
Equating the two expressions for efficiency: $0.2 = 1 - \frac{400}{T_{1}}$.
$\frac{400}{T_{1}} = 1 - 0.2 = 0.8$.
$T_{1} = \frac{400}{0.8} = 500 \, K$.
Thus,the minimum temperature of the hot reservoir must be $500 \, K$.

(A) Let $T_1$ be the temperature of the source and $T_2$ be the temperature of the sink in Kelvin.
Efficiency $\eta = 1 - \frac{T_2}{T_1}$.
Given $\eta = \frac{1}{4}$,so $\frac{1}{4} = 1 - \frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1} = \frac{3}{4} \dots (i)$.
When the sink temperature is reduced by $58^{\circ}C$ (which is equivalent to $58 \ K$ in temperature difference),the new sink temperature is $T_2' = T_2 - 58$.
The new efficiency is double,so $\eta' = 2 \times \frac{1}{4} = \frac{1}{2}$.
Thus,$\frac{1}{2} = 1 - \frac{T_2 - 58}{T_1} \Rightarrow \frac{T_2 - 58}{T_1} = \frac{1}{2}$.
Substituting $\frac{T_2}{T_1} = \frac{3}{4}$ into the equation: $\frac{3}{4} - \frac{58}{T_1} = \frac{1}{2}$.
$\frac{58}{T_1} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \Rightarrow T_1 = 58 \times 4 = 232 \ K$.
Now,$T_2 = \frac{3}{4} \times 232 = 174 \ K$.
Since the temperature in Kelvin is $174 \ K$,we convert it to Celsius: $T_C = 174 - 273.15 = -99.15^{\circ}C$. However,assuming the question implies the temperature values are in Kelvin and the result is requested in Celsius based on standard textbook problem conventions where $T(K) = T(^{\circ}C) + 273$,the sink temperature is $174 \ K$,which is $-99^{\circ}C$. Given the options,there is a discrepancy in the unit interpretation. Re-evaluating: if $T_2 = 174 \ K$,the answer is $174$ if the question implies the value in Kelvin.

(B) 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 in Kelvin.
In the first case,$\eta_1 = \frac{1}{6} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{5}{6}$ (Equation $i$).
In the second case,the sink temperature is reduced by $62^{\circ}C$ (which is equivalent to $62 K$ in temperature difference),so the new sink temperature is $T_2' = T_2 - 62$. The new efficiency is $\eta_2 = 2 \times \frac{1}{6} = \frac{1}{3}$.
Thus,$\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$,which implies $\frac{T_2 - 62}{T_1} = \frac{2}{3}$ (Equation $ii$).
Subtracting Equation $ii$ from Equation $i$: $\frac{T_2}{T_1} - \frac{T_2 - 62}{T_1} = \frac{5}{6} - \frac{2}{3}$.
$\frac{62}{T_1} = \frac{5-4}{6} = \frac{1}{6}$.
Therefore,$T_1 = 62 \times 6 = 372 K$.
To convert Kelvin to Celsius: $T(^{\circ}C) = 372 - 273 = 99^{\circ}C$.

(A) For Carnot engine $A$: Efficiency $\eta_A = 1 - \frac{T}{T_1} = \frac{W_A}{Q_1}$. Thus,$W_A = Q_1 \left(1 - \frac{T}{T_1}\right) = Q_1 - \frac{Q_1 T}{T_1}$.
Since $\frac{Q}{Q_1} = \frac{T}{T_1}$,we have $Q = \frac{Q_1 T}{T_1}$.
For Carnot engine $B$: It absorbs heat $Q_B = Q/2$ at temperature $T$ and rejects heat $Q_3$ at temperature $T_3$. Efficiency $\eta_B = 1 - \frac{T_3}{T} = \frac{W_B}{Q_B}$.
Thus,$W_B = \frac{Q}{2} \left(1 - \frac{T_3}{T}\right) = \frac{Q}{2} - \frac{Q T_3}{2 T}$.
Given $W_A = W_B$,so $Q_1 - Q = \frac{Q}{2} - Q_3$.
Substituting $Q = \frac{Q_1 T}{T_1}$,we get $Q_1 - \frac{Q_1 T}{T_1} = \frac{Q_1 T}{2 T_1} - Q_3$.
Rearranging terms: $Q_1 + Q_3 = \frac{Q_1 T}{T_1} + \frac{Q_1 T}{2 T_1} = \frac{3 Q_1 T}{2 T_1}$.
Using the relation for engine $B$,$\frac{Q_3}{Q/2} = \frac{T_3}{T} \Rightarrow Q_3 = \frac{Q T_3}{2 T} = \frac{Q_1 T T_3}{2 T_1 T} = \frac{Q_1 T_3}{2 T_1}$.
Substituting $Q_3$ back: $Q_1 + \frac{Q_1 T_3}{2 T_1} = \frac{3 Q_1 T}{2 T_1}$.
Dividing by $Q_1$: $1 + \frac{T_3}{2 T_1} = \frac{3 T}{2 T_1}$.
Multiplying by $\frac{2 T_1}{3}$: $T = \frac{2 T_1}{3} + \frac{T_3}{3} = \frac{2}{3} T_1 + \frac{1}{3} T_3$.

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.
Given: $Q_H = 5000 \, kcal$,$T_H = 727 + 273 = 1000 \, K$,$T_L = 127 + 273 = 400 \, K$.
Efficiency $\eta = 1 - \frac{400}{1000} = 1 - 0.4 = 0.6$.
Work done $W = \eta \times Q_H = 0.6 \times 5000 \, kcal = 3000 \, kcal$.
Since $1 \, kcal = 4184 \, J$,then $W = 3000 \times 4184 \, J = 12,552,000 \, J$.
Rounding to the nearest value,$W \approx 12.6 \times 10^{6} \, J$.

(B) The efficiency of a Carnot engine 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 $T_L = 27\,^{\circ} C = 300\, K$ and $\eta = 25\% = 0.25$.
Substituting the values: $0.25 = 1 - \frac{300}{T_H} \implies \frac{300}{T_H} = 0.75 \implies T_H = \frac{300}{0.75} = 400\, K$.
If the efficiency is increased by $100\%$ of its original value,the new efficiency $\eta' = \eta + 100\% \text{ of } \eta = 0.25 + 0.25 = 0.50$ or $50\%$.
Let the new source temperature be $T_H'$.
$0.50 = 1 - \frac{300}{T_H'} \implies \frac{300}{T_H'} = 0.50 \implies T_H' = \frac{300}{0.50} = 600\, K$.
The change in temperature is $\Delta T = T_H' - T_H = 600\, K - 400\, K = 200\, K$.
Since a change in temperature of $1\, K$ is equivalent to a change of $1\,^{\circ} C$,the temperature should be increased by $200\,^{\circ} C$.

(B) For a reversible (Carnot) heat engine,the efficiency is given by $\eta = 1 - \frac{Q_L}{Q_H} = 1 - \frac{T_L}{T_H}$.
Given:
Temperature of the cold reservoir,$T_L = 324 \; K$.
Heat taken from the hot reservoir,$Q_H = 300 \; J$.
Heat delivered to the cold reservoir,$Q_L = 180 \; J$.
Using the relation for a reversible engine: $\frac{Q_L}{Q_H} = \frac{T_L}{T_H}$.
Substituting the values:
$\frac{180}{300} = \frac{324}{T_H}$.
$T_H = \frac{324 \times 300}{180}$.
$T_H = \frac{324 \times 5}{3} = 108 \times 5 = 540 \; K$.
Thus,the minimum temperature of the hot reservoir is $540 \; K$.

(A) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = \left(1 - \frac{T_L}{T_H}\right) \times 100\,\%$
Here,the ice point is $T_L = 0\,^{\circ}C = 273\,K$ and the steam point is $T_H = 100\,^{\circ}C = 373\,K$.
Substituting these values into the formula:
$\eta = \left(1 - \frac{273}{373}\right) \times 100\,\%$
$\eta = \left(\frac{373 - 273}{373}\right) \times 100\,\%$
$\eta = \left(\frac{100}{373}\right) \times 100\,\%$
$\eta \approx 0.2681 \times 100\,\% = 26.81\,\%$

(A) Given:
Temperature of the reservoir,$T_{1} = 527^{\circ} C = 527 + 273 = 800 \; K$.
Temperature of the sink,$T_{2} = 200 \; K$.
Work done,$W = 12000 \; kJ = 12 \times 10^{6} \; J$.
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 absorbed from the reservoir.
Equating the two expressions for efficiency:
$1 - \frac{200}{800} = \frac{12 \times 10^{6}}{Q_{1}}$
$1 - \frac{1}{4} = \frac{12 \times 10^{6}}{Q_{1}}$
$\frac{3}{4} = \frac{12 \times 10^{6}}{Q_{1}}$
$Q_{1} = \frac{12 \times 10^{6} \times 4}{3}$
$Q_{1} = 16 \times 10^{6} \; J$.
Thus,the value of $x$ is $16$.

(D) For a reversible heat engine,the efficiency is given by $\eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$.
Given $Q_1 = 300 \, cal$ and $Q_2 = 225 \, cal$.
The source temperature $T_1 = 227^{\circ} C = 227 + 273 = 500 \, K$.
Using the relation $\frac{Q_2}{Q_1} = \frac{T_2}{T_1}$:
$\frac{225}{300} = \frac{T_2}{500}$.
$T_2 = \frac{225 \times 500}{300} = \frac{225 \times 5}{3} = 75 \times 5 = 375 \, K$.
To convert the sink temperature to Celsius: $T_2(^{\circ} C) = 375 - 273 = 102^{\circ} C$.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{L}}{T_{H}}$,where temperatures must be in Kelvin $(K = ^{\circ}C + 273)$.
For $\eta_{1}$:
$T_{H} = 447 + 273 = 720 \ K$
$T_{L} = 147 + 273 = 420 \ K$
$\eta_{1} = 1 - \frac{420}{720} = 1 - \frac{42}{72} = 1 - \frac{7}{12} = \frac{5}{12}$.
For $\eta_{2}$:
$T_{H} = 947 + 273 = 1220 \ K$
$T_{L} = 47 + 273 = 320 \ K$
$\eta_{2} = 1 - \frac{320}{1220} = 1 - \frac{32}{122} = 1 - \frac{16}{61} = \frac{45}{61}$.
Now,the ratio $\frac{\eta_{1}}{\eta_{2}}$ is:
$\frac{\eta_{1}}{\eta_{2}} = \frac{5/12}{45/61} = \frac{5}{12} \times \frac{61}{45} = \frac{61}{12 \times 9} = \frac{61}{108} \approx 0.5648$.
Rounding to two decimal places,the ratio is $0.56$.

(A) First case: The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.
For the first case,$\eta = 1 - \frac{100}{300} = 1 - \frac{1}{3} = \frac{2}{3}$.
Second case: Two engines are connected in series. The net efficiency $\eta_{\text{net}}$ of two engines in series is given by $\eta_{\text{net}} = \eta_1 + \eta_2 - \eta_1 \eta_2$.
For the first engine $(E_1)$ operating between $300 \, K$ and $200 \, K$,$\eta_1 = 1 - \frac{200}{300} = 1 - \frac{2}{3} = \frac{1}{3}$.
For the second engine $(E_2)$ operating between $200 \, K$ and $100 \, K$,$\eta_2 = 1 - \frac{100}{200} = 1 - \frac{1}{2} = \frac{1}{2}$.
Now,calculating the net efficiency: $\eta_{\text{net}} = \frac{1}{3} + \frac{1}{2} - (\frac{1}{3} \times \frac{1}{2}) = \frac{1}{3} + \frac{1}{2} - \frac{1}{6} = \frac{2+3-1}{6} = \frac{4}{6} = \frac{2}{3}$.
Thus,the efficiency in the $2^{\text{nd}}$ case is the same as in the $1^{\text{st}}$ case.

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.
Given $\eta_1 = 0.5$,so $0.5 = 1 - \frac{T_L}{T_H} \implies \frac{T_L}{T_H} = 0.5 \implies T_L = 0.5 T_H$.
When the sink temperature is reduced by $40 \, K$,the new sink temperature is $T_L' = T_L - 40$.
The new efficiency $\eta_2$ increases by $30 \%$ of the original efficiency,so $\eta_2 = 0.5 + 0.3 = 0.8$.
Using the formula for the new efficiency: $0.8 = 1 - \frac{T_L - 40}{T_H}$.
Substituting $T_L = 0.5 T_H$: $0.8 = 1 - \frac{0.5 T_H - 40}{T_H}$.
$0.8 = 1 - 0.5 + \frac{40}{T_H}$.
$0.8 = 0.5 + \frac{40}{T_H} \implies 0.3 = \frac{40}{T_H}$.
$T_H = \frac{40}{0.3} = 133.33 \, K$.
Wait,checking the calculation: $0.5(1.3) = 0.65$. $0.65 = 1 - \frac{T_L - 40}{T_H} \implies \frac{T_L - 40}{T_H} = 0.35$.
Since $\frac{T_L}{T_H} = 0.5$,then $0.5 - \frac{40}{T_H} = 0.35 \implies \frac{40}{T_H} = 0.15 \implies T_H = \frac{40}{0.15} = 266.67 \, K$.

(D) The given cycle is a Carnot cycle consisting of two isothermal and two adiabatic processes.
The efficiency $\eta$ of a Carnot cycle is given by $\eta = 1 - \frac{T_{L}}{T_{H}}$, where $T_{L}$ is the temperature of the cold reservoir and $T_{H}$ is the temperature of the hot reservoir.
For an adiabatic process, $TV^{\gamma-1} = \text{constant}$.
In the adiabatic compression step $(A \to B)$, the gas is compressed from $V_{1}$ to $V_{B} = 1 \; m^{3}$ at temperature $T_{L}$ to $T_{H}$.
Thus, $T_{L} V_{1}^{\gamma-1} = T_{H} V_{B}^{\gamma-1}$.
$\frac{T_{L}}{T_{H}} = \left( \frac{V_{B}}{V_{1}} \right)^{\gamma-1} = \left( \frac{1}{V_{1}} \right)^{\gamma-1}$.
Given $\eta = 3/4$, so $1 - \frac{T_{L}}{T_{H}} = 3/4$, which implies $\frac{T_{L}}{T_{H}} = 1/4$.
For a monoatomic gas, the adiabatic index $\gamma = 5/3$, so $\gamma - 1 = 2/3$.
Substituting these values: $\left( \frac{1}{V_{1}} \right)^{2/3} = 1/4$.
Taking the reciprocal: $V_{1}^{2/3} = 4$.
Raising both sides to the power of $3/2$: $V_{1} = 4^{3/2} = (2^{2})^{3/2} = 2^{3} = 8 \; m^{3}$.

(A) The efficiency $\eta$ of a heat engine is defined as the ratio of the work done $W$ by the engine to the heat $Q_H$ absorbed from the source.
$\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}$
According to the second law of thermodynamics,no heat engine can have an efficiency of $1$ $(100\%)$,as it would require the heat rejected to the sink $Q_L$ to be $0$,which is impossible.
Also,the efficiency cannot be negative as it would imply the engine is performing work by absorbing heat from a cold reservoir without external work,violating the second law.
Therefore,the efficiency $\eta$ always lies between $0$ and $1$ $(0 < \eta < 1)$.

(C) The efficiency of a Carnot engine is given by the formula $\eta = 1 - \frac{T_1}{T_2}$,where $T_1$ is the temperature of the sink and $T_2$ is the temperature of the source in Kelvin.
Given:
$T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$
$T_2 = 527^{\circ} C = 527 + 273 = 800 \ K$
Substituting the values into the formula:
$\eta = 1 - \frac{400}{800}$
$\eta = 1 - 0.5 = 0.5$
To express the efficiency as a percentage:
$\text{Efficiency} = 0.5 \times 100 = 50 \%$

(B) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.
Here,$T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.
Given that the temperature of the sink is at absolute zero,we have $T_2 = 0 \ K$.
Substituting this value into the efficiency formula:
$\eta = 1 - \frac{0}{T_1} = 1 - 0 = 1$.
To express the efficiency as a percentage,we multiply by $100$:
$\eta = 1 \times 100\% = 100\%$.
Therefore,the efficiency of the Carnot engine is $100\%$.

(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 $T_2 = 300 \, K$ and initial efficiency $\eta = 0.5$.
$0.5 = 1 - \frac{300}{T_1} \Rightarrow \frac{300}{T_1} = 0.5 \Rightarrow T_1 = 600 \, K$.
Now,the new efficiency $\eta' = 0.7$ with the same sink temperature $T_2 = 300 \, K$.
$0.7 = 1 - \frac{300}{T_1'} \Rightarrow \frac{300}{T_1'} = 0.3 \Rightarrow T_1' = \frac{300}{0.3} = 1000 \, K$.
The increase in source temperature is $\Delta T = T_1' - T_1 = 1000 \, K - 600 \, K = 400 \, K$.

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
Given temperatures are $T_1 = 627^{\circ} C = 627 + 273 = 900 \, K$ and $T_2 = 27^{\circ} C = 27 + 273 = 300 \, K$.
The heat taken from the source is $Q_1 = 6000 \, cal = 6 \, kcal$.
The efficiency is $\eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}$.
Since $\eta = \frac{W}{Q_1}$,the work done $W = \eta \times Q_1$.
$W = \frac{2}{3} \times 6 \, kcal = 4 \, kcal$.

(B) The correct answer is $(B)$.
$A$ Carnot engine operates on a reversible cycle,which is an idealized theoretical model.
In the real world,all thermodynamic processes involve friction,turbulence,and heat loss to the surroundings,making them inherently irreversible.
Since a perfectly reversible process cannot be achieved in practice,real heat engines always have lower efficiency than the theoretical Carnot engine operating between the same two temperatures.

(B) The correct answer is $B$.
According to the Carnot theorem,the efficiency of a reversible engine operating between two given temperatures is the maximum possible efficiency for any engine operating between those same temperatures.
An irreversible engine involves dissipative effects like friction,turbulence,and heat loss,which reduce the amount of heat converted into work.
Therefore,the efficiency of a reversible engine is always greater than that of an irreversible engine operating between the same two heat reservoirs.

(B) The correct answer is $B$.
The Carnot cycle is one of the foundations of the second law of thermodynamics,and Sadi Carnot is often considered the father of thermodynamics. He was one of the pioneers who first determined an idealistic way of converting heat energy into work. The Carnot cycle represents the most efficient heat engine cycle.
The Carnot cycle consists of the following four processes:
$I.$ Isothermal expansion: The gas undergoes an isothermal expansion at a high temperature $T_{\text{high}}$. In this process,the gas absorbs $q_{\text{in}}$ amount of heat from the source and performs work $w_1$ on the surroundings.
$II.$ Reversible adiabatic expansion: The gas then undergoes a reversible adiabatic expansion. Consequently,the temperature of the gas decreases to a lower temperature $T_{\text{low}}$.
$III.$ Isothermal compression: The gas is then compressed isothermally at the lower temperature $T_{\text{low}}$. In this process,the gas releases $q_{\text{out}}$ amount of heat to the sink,and work is done on the gas by the surroundings.
$IV.$ Reversible adiabatic compression: Finally,the gas undergoes a reversible adiabatic compression,which causes the temperature to rise back to $T_{\text{high}}$.
Thus,the Carnot cycle consists of two isothermal and two adiabatic processes,totaling four stages.

(D) The efficiency of a heat engine is defined by the Carnot theorem,which states that the maximum possible efficiency for any heat engine operating between two temperatures is the efficiency of a Carnot engine.
The formula for the efficiency $(\eta)$ of a Carnot engine is given by:
$\eta = 1 - \frac{T_2}{T_1}$
Where:
$T_1$ is the absolute temperature of the source.
$T_2$ is the absolute temperature of the sink.
Thus,the maximum possible efficiency depends on the temperatures of the source and the sink.