Heat Engine and Carnot Cycle Questions in English

(D) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = \frac{T_1 - T_2}{T_1} = \frac{W}{Q}$,where $T_1$ is the source temperature,$T_2$ is the sink temperature,$W$ is the work output,and $Q$ is the heat supplied.
Given: $T_1 = 600\,K$,$T_2 = 300\,K$,and $W = 800\,J$.
Rearranging the formula to solve for $Q$: $Q = \left( \frac{T_1}{T_1 - T_2} \right) W$.
Substituting the values: $Q = \left( \frac{600}{600 - 300} \right) \times 800$.
$Q = \left( \frac{600}{300} \right) \times 800 = 2 \times 800 = 1600\,J$.
Thus,the heat energy supplied per cycle is $1600\,J$.

(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ and $T_2$ are the absolute temperatures in Kelvin.
Given $T_1 = 227^{\circ}C = 227 + 273 = 500 \, K$ and $T_2 = 127^{\circ}C = 127 + 273 = 400 \, K$.
$\eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2$ or $\frac{1}{5}$.
Since efficiency $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat absorbed at high temperature:
$W = \eta \times Q_1 = \frac{1}{5} \times (6 \times 10^4 \, J) = 1.2 \times 10^4 \, J$.

(B) 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 in Kelvin.
Given: $\eta = 30\% = 0.3$,$T_2 = 77^{\circ}C = 77 + 273 = 350\,K$.
Substituting the values: $0.3 = 1 - \frac{350}{T_1}$.
Rearranging the equation: $\frac{350}{T_1} = 1 - 0.3 = 0.7$.
Solving for $T_1$: $T_1 = \frac{350}{0.7} = 500\,K$.
Converting back to Celsius: $T_1 = 500 - 273 = 227^{\circ}C$.

(B) 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.
For the efficiency to be $100\%$,we set $\eta = 1$.
Substituting this into the formula: $1 = 1 - \frac{T_2}{T_1}$.
This simplifies to $\frac{T_2}{T_1} = 0$,which implies $T_2 = 0\, K$ (absolute zero).
Therefore,the efficiency of a Carnot engine is $100\%$ only if the sink temperature is at absolute zero.

(C) The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.
Here,the source temperature $T_1 = 411 + 273 = 684 \, K$ and the sink temperature $T_2 = 69 + 273 = 342 \, K$.
Substituting these values into the efficiency formula: $\eta = 1 - \frac{342}{684} = 1 - 0.5 = 0.5$.
The work done per cycle is equal to the area enclosed by the $PV$ diagram,which is given by $W = \eta \times Q$.
Given that the heat extracted $Q = 1000 \, J$,we have $W = 0.5 \times 1000 = 500 \, J$.

(B) For a Carnot engine,the efficiency $\eta$ is given by the ratio of temperatures of the sink $(T_2)$ and the source $(T_1)$:
$\eta = 1 - \frac{T_2}{T_1}$
Given,$T_1 = T$ and $T_2 = T/3$.
Therefore,$\eta = 1 - \frac{T/3}{T} = 1 - \frac{1}{3} = \frac{2}{3}$.
Also,the efficiency is defined as $\eta = 1 - \frac{Q_2}{Q_1}$,where $Q_1$ is the heat absorbed and $Q_2$ is the heat rejected.
Equating the two expressions for efficiency:
$\frac{2}{3} = 1 - \frac{Q_2}{Q}$
$\frac{Q_2}{Q} = 1 - \frac{2}{3} = \frac{1}{3}$
$Q_2 = \frac{Q}{3}$.
Thus,the amount of heat rejected is $Q/3$.

(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 in Kelvin.
Given: Sink temperature $T_2 = 27^\circ C = 27 + 273 = 300 \ K$.
Efficiency $\eta = 25\% = 0.25$.
Substituting the values into the formula: $0.25 = 1 - \frac{300}{T_1}$.
Rearranging the terms: $\frac{300}{T_1} = 1 - 0.25 = 0.75$.
$T_1 = \frac{300}{0.75} = 400 \ K$.
Converting back to Celsius: $T_1 = 400 - 273 = 127^\circ C$.

(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.
Given: $\eta = 70\% = 0.7$ and $T_1 = 1000 \ K$.
Substituting the values into the formula: $0.7 = 1 - \frac{T_2}{1000}$.
Rearranging the terms: $\frac{T_2}{1000} = 1 - 0.7 = 0.3$.
Therefore,$T_2 = 0.3 \times 1000 = 300 \ K$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ and $T_2$ are the absolute temperatures in Kelvin.
First,convert the temperatures to Kelvin: $T(K) = T(^oC) + 273.15$. For calculation,we use $273$.
For the first case: $T_1 = 200 + 273 = 473 \ K$ and $T_2 = 0 + 273 = 273 \ K$.
$\eta_1 = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$.
For the second case: $T_1 = 0 + 273 = 273 \ K$ and $T_2 = -200 + 273 = 73 \ K$.
$\eta_2 = 1 - \frac{73}{273} = \frac{273 - 73}{273} = \frac{200}{273}$.
Now,calculate the ratio: $\frac{\eta_1}{\eta_2} = \frac{200/473}{200/273} = \frac{273}{473} \approx 0.577$.

(A) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_{sink}}{T_{source}}$.
First, convert the temperatures from Celsius to Kelvin:
$T_{source} = 27^{\circ}C = 27 + 273 = 300 \ K$.
$T_{sink} = -123^{\circ}C = -123 + 273 = 150 \ K$.
Now, substitute these values into the efficiency formula:
$\eta = 1 - \frac{150}{300} = 1 - 0.5 = 0.5$.
To express this as a percentage, multiply by $100$:
$\eta = 0.5 \times 100 = 50\%$.

(B) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink in Kelvin.
First,convert the temperatures from Celsius to Kelvin:
$T_1 = 227 + 273 = 500 \ K$
$T_2 = 27 + 273 = 300 \ K$
Now,substitute these values into the efficiency formula:
$\eta = 1 - \frac{300}{500}$
$\eta = 1 - 0.6 = 0.4$
Therefore,the efficiency of the engine is $0.4$.

(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 the first case,the temperatures are $T_H = 800 \ K$ and $T_L = 500 \ K$. Thus,$\eta_1 = 1 - \frac{500}{800} = 1 - \frac{5}{8} = \frac{3}{8}$.
In the second case,the temperatures are $T_H = x \ K$ and $T_L = 600 \ K$. Thus,$\eta_2 = 1 - \frac{600}{x}$.
Since the efficiencies are equal,$\eta_1 = \eta_2$,we have $\frac{3}{8} = 1 - \frac{600}{x}$.
Rearranging the terms,$\frac{600}{x} = 1 - \frac{3}{8} = \frac{5}{8}$.
Solving for $x$,we get $x = \frac{600 \times 8}{5} = 120 \times 8 = 960 \ K$.

(A) The maximum theoretical efficiency of a heat engine is given by the Carnot efficiency formula: $\eta_{\max} = 1 - \frac{T_2}{T_1}$.
First,convert the temperatures from Celsius to Kelvin: $T_1 = 127 + 273 = 400 \text{ K}$ and $T_2 = 27 + 273 = 300 \text{ K}$.
Substituting these values into the formula: $\eta_{\max} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$ or $25\%$.
Since the second law of thermodynamics states that no heat engine can have an efficiency greater than the Carnot efficiency operating between the same two temperatures,an efficiency of $26\%$ is impossible.

(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 in Kelvin.
For the first case: $T_H = 200 + 273 = 473 \ K$ and $T_L = 0 + 273 = 273 \ K$.
$\eta_1 = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$.
For the second case: $T_H = 0 + 273 = 273 \ K$ and $T_L = -200 + 273 = 73 \ K$.
$\eta_2 = 1 - \frac{73}{273} = \frac{273 - 73}{273} = \frac{200}{273}$.
The ratio of efficiencies is $\frac{\eta_1}{\eta_2} = \frac{200/473}{200/273} = \frac{273}{473} \approx \frac{1}{1.73}$.
Thus,the ratio is $1 : 1.73$.

(B) 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.
Initially,$\eta_1 = 0.5$ and $T_2 = 500 \ K$.
$0.5 = 1 - \frac{500}{T_1} \implies \frac{500}{T_1} = 0.5 \implies T_1 = 1000 \ K$.
Now,the efficiency is increased to $\eta_2 = 0.6$ while keeping $T_1$ constant at $1000 \ K$.
$0.6 = 1 - \frac{T_2'}{1000} \implies \frac{T_2'}{1000} = 1 - 0.6 = 0.4$.
$T_2' = 0.4 \times 1000 = 400 \ K$.
Therefore,the required temperature of the sink is $400 \ K$.

(C) The efficiency 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.
To obtain $100\%$ efficiency,we require $\eta = 1$,which implies $\frac{T_2}{T_1} = 0$.
This condition can only be met if $T_2 = 0 \text{ K}$ (absolute zero temperature).
According to the $3\text{rd}$ law of thermodynamics,it is impossible to reach absolute zero temperature in a finite number of steps.
Furthermore,if $T_2$ were $0 \text{ K}$,all heat taken from the source would be converted into work,which violates the $2\text{nd}$ law of thermodynamics (Kelvin-Planck statement).
Therefore,we cannot reach absolute zero temperature.

(B) The efficiency 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: $T_1 = 627^{\circ}C = 627 + 273 = 900 \, \text{K}$,$T_2 = 27^{\circ}C = 27 + 273 = 300 \, \text{K}$,and $Q_1 = 3 \times 10^6 \, \text{cal}$.
Converting heat to Joules: $Q_1 = 3 \times 10^6 \times 4.2 \, \text{J} = 12.6 \times 10^6 \, \text{J}$.
The efficiency is $\eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}$.
Since $\eta = \frac{W}{Q_1}$,the work done is $W = \eta \times Q_1 = \frac{2}{3} \times 12.6 \times 10^6 \, \text{J} = 2 \times 4.2 \times 10^6 \, \text{J} = 8.4 \times 10^6 \, \text{J}$.

(D) 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.
To maximize the efficiency $\eta$,the ratio $\frac{T_2}{T_1}$ must be minimized.
Let us calculate the ratio for each option:
$(a)$ $\frac{60}{80} = 0.75$
$(b)$ $\frac{80}{100} = 0.80$
$(c)$ $\frac{40}{60} \approx 0.667$
$(d)$ $\frac{20}{40} = 0.50$
Comparing these values,the ratio is minimum for option $(d)$. Therefore,the efficiency is highest for the combination $40 K$ and $20 K$.

(A) The efficiency $(\eta)$ of a Carnot engine is defined as the ratio of the work done by the engine to the heat absorbed from the source.
For a reversible Carnot cycle,the efficiency depends only on the temperatures of the source and the sink.
The formula for 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 absolute temperature of the source (in Kelvin).
$T_2$ is the absolute temperature of the sink (in Kelvin).
Therefore,the correct option is $A$.

(B) The efficiency of an ideal heat engine (Carnot engine) is given by the formula: $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$.
In the first case,the efficiency is $\eta = \frac{T_1 - T_2}{T_1}$.
In the second case,the new source temperature is $T_1' = 2T_1$ and the new sink temperature is $T_2' = 2T_2$.
The new efficiency $\eta'$ is given by: $\eta' = \frac{T_1' - T_2'}{T_1'} = \frac{2T_1 - 2T_2}{2T_1}$.
Simplifying the expression: $\eta' = \frac{2(T_1 - T_2)}{2T_1} = \frac{T_1 - T_2}{T_1} = \eta$.
Therefore,the efficiency remains unchanged.

(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = \frac{T_1 - T_2}{T_1} = \frac{W}{Q_1}$.
Here,$T_1 = 227^{\circ}C = 227 + 273 = 500 \text{ K}$ and $T_2 = 127^{\circ}C = 127 + 273 = 400 \text{ K}$.
The heat absorbed at the higher temperature is $Q_1 = 6 \times 10^4 \text{ cal}$.
Using the formula $W = Q_1 \left( \frac{T_1 - T_2}{T_1} \right)$:
$W = 6 \times 10^4 \times \left( \frac{500 - 400}{500} \right)$
$W = 6 \times 10^4 \times \left( \frac{100}{500} \right)$
$W = 6 \times 10^4 \times 0.2 = 1.2 \times 10^4 \text{ cal}$.
Thus,the amount of heat converted to work is $1.2 \times 10^4 \text{ cal}$.

(C) For Carnot engine $A$,the efficiency is $\eta_A = \frac{T_1 - T_2}{T_1} = \frac{W_A}{Q_1}$.
For Carnot engine $B$,the efficiency is $\eta_B = \frac{T_2 - T_3}{T_2} = \frac{W_B}{Q_2}$.
Since the heat rejected by engine $A$ is the heat received by engine $B$,we have $Q_2 = Q_1(1 - \eta_A) = Q_1 \left( \frac{T_2}{T_1} \right)$.
Given that the work outputs are equal,$W_A = W_B$,which implies $Q_1 \eta_A = Q_2 \eta_B$.
Substituting the expressions: $Q_1 \left( \frac{T_1 - T_2}{T_1} \right) = Q_1 \left( \frac{T_2}{T_1} \right) \left( \frac{T_2 - T_3}{T_2} \right)$.
Simplifying this,we get $T_1 - T_2 = T_2 - T_3$.
Therefore,$2T_2 = T_1 + T_3$,which gives $T_2 = \frac{T_1 + T_3}{2}$.
Substituting the given values: $T_2 = \frac{800 + 300}{2} = \frac{1100}{2} = 550 \ K$.

(D) The efficiency of a reversible 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.
Initially,$\eta = \frac{1}{6}$,so $1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$ ...$(i)$
When the sink temperature is reduced by $62^\circ C$ (which is equivalent to $62 \ K$),the new efficiency $\eta' = 2\eta = 2 \times \frac{1}{6} = \frac{1}{3}$.
The new sink temperature is $T_2' = T_2 - 62$.
Thus,$\eta' = 1 - \frac{T_2 - 62}{T_1} = \frac{1}{3}$.
Substituting $T_2 = \frac{5}{6}T_1$ into the equation: $1 - \frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{1}{3}$.
$1 - (\frac{5}{6} - \frac{62}{T_1}) = \frac{1}{3} \implies 1 - \frac{5}{6} + \frac{62}{T_1} = \frac{1}{3}$.
$\frac{1}{6} + \frac{62}{T_1} = \frac{1}{3} \implies \frac{62}{T_1} = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$.
$T_1 = 62 \times 6 = 372 \ K = (372 - 273)^\circ C = 99^\circ C$.
Now,$T_2 = \frac{5}{6} \times 372 = 310 \ K = (310 - 273)^\circ C = 37^\circ C$.

(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.
Given $T_2 = 7 + 273 = 280 \, K$ and initial efficiency $\eta_1 = 0.5$.
$0.5 = 1 - \frac{280}{T_1} \implies \frac{280}{T_1} = 0.5 \implies T_1 = 560 \, K$.
For the new efficiency $\eta_2 = 0.7$ with the same cold reservoir temperature $T_2 = 280 \, K$:
$0.7 = 1 - \frac{280}{T_1'} \implies \frac{280}{T_1'} = 0.3 \implies T_1' = \frac{280}{0.3} = 933.33 \, K$.
The increase in temperature of the high temperature reservoir is $\Delta T = T_1' - T_1 = 933.33 - 560 = 373.33 \, K$.
Rounding to the nearest provided option,the increase is approximately $380 \, K$.

(D) $1$. The area enclosed by a closed loop in a $P-V$ diagram represents the net work done by the gas during one complete cycle. Since the cycle is clockwise,the work done is positive (work done by the gas). Thus,statement $I$ is correct.
$2$. According to the first law of thermodynamics,$\Delta Q = \Delta W + \Delta U$. For a complete cycle,the change in internal energy $\Delta U$ is zero because internal energy is a state function and the system returns to its initial state. Thus,statement $III$ is correct.
$3$. Since $\Delta U = 0$,the first law becomes $\Delta Q = \Delta W$. This means the net heat absorbed by the gas is equal to the net work done by the gas,which is equal to the area enclosed by the cycle $ABCD$. Thus,statement $II$ is correct.
Therefore,statements $I, II,$ and $III$ are all correct.

(A) The area under the $T-S$ curve represents the heat exchanged in a process.
For the given cycle, the heat absorbed $(Q_{in})$ is the area under the top slanted line from $S_0$ to $2S_0$.
The area is a rectangle plus a triangle: $Q_{in} = (T_0 \times S_0) + \frac{1}{2} \times (2T_0 - T_0) \times (2S_0 - S_0) = T_0 S_0 + \frac{1}{2} T_0 S_0 = 1.5 T_0 S_0$.
The heat rejected $(Q_{out})$ is the area under the bottom horizontal line from $2S_0$ to $S_0$ at temperature $T_0$: $Q_{out} = T_0 \times (2S_0 - S_0) = T_0 S_0$.
The efficiency $\eta$ is given by $\eta = 1 - \frac{Q_{out}}{Q_{in}}$.
Substituting the values: $\eta = 1 - \frac{T_0 S_0}{1.5 T_0 S_0} = 1 - \frac{1}{1.5} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$.

(B) The temperatures are $T_{1} = 627 + 273 = 900 \text{ K}$ and $T_{2} = 27 + 273 = 300 \text{ K}$.
Efficiency $\eta = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}$.
Heat absorbed $Q_{1} = 3 \times 10^{6} \text{ cal} = 3 \times 10^{6} \times 4.2 \text{ J} = 12.6 \times 10^{6} \text{ J}$.
Work done $W = \eta \times Q_{1} = \frac{2}{3} \times 12.6 \times 10^{6} \text{ J} = 8.4 \times 10^{6} \text{ J}$.

(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.
To maximize $\eta$,the ratio $\frac{T_2}{T_1}$ must be minimized.
Calculating the ratio for each option:
$A: \frac{60}{80} = 0.75$
$B: \frac{80}{100} = 0.80$
$C: \frac{40}{60} \approx 0.67$
$D: \frac{20}{40} = 0.50$
Since the ratio is minimum for option $D$,the efficiency is highest for $T_1 = 40 \ K$ and $T_2 = 20 \ K$.

(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 in Kelvin.
Given: $T_1 = 227 + 273 = 500 \, K$ and $T_2 = 127 + 273 = 400 \, K$.
Heat absorbed from the source $Q_1 = 6 \times 10^{4} \, cal$.
The efficiency is also defined as $\eta = \frac{W}{Q_1}$,where $W$ is the net work done.
Therefore,$W = Q_1 \left( 1 - \frac{T_2}{T_1} \right)$.
Substituting the values: $W = 6 \times 10^{4} \left( 1 - \frac{400}{500} \right)$.
$W = 6 \times 10^{4} \left( 1 - 0.8 \right) = 6 \times 10^{4} \times 0.2$.
$W = 1.2 \times 10^{4} \, cal$.

(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 source temperature and $T_2$ is the sink temperature in Kelvin.
Convert temperatures from Celsius to Kelvin: $T_1 = 27 + 273 = 300 \, K$ and $T_2 = -123 + 273 = 150 \, K$.
Substitute the values into the formula: $\eta = 1 - \frac{150}{300}$.
Simplify the expression: $\eta = 1 - 0.5 = 0.5$.
Convert to percentage: $\eta = 0.5 \times 100 = 50\%$.

(A) Given: Heat supplied by the source $Q_1 = 200 \, cal$,Heat rejected to the sink $Q_2 = 150 \, cal$,Temperature of the source $T_1 = 400 \, K$.
For a reversible heat engine,the ratio of heat exchanged is equal to the ratio of temperatures: $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$.
Rearranging the formula to solve for the sink temperature $T_2$: $T_2 = \frac{Q_2}{Q_1} \times T_1$.
Substituting the given values: $T_2 = \frac{150}{200} \times 400$.
$T_2 = 0.75 \times 400 = 300 \, K$.
Therefore,the temperature of the sink is $300 \, K$.

(C) The efficiency 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 in Kelvin.
Given: $\eta = 25\% = 0.25$ and $T_2 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Substituting the values: $0.25 = 1 - \frac{300}{T_1}$.
Rearranging the terms: $\frac{300}{T_1} = 1 - 0.25 = 0.75$.
$T_1 = \frac{300}{0.75} = 400 \ K$.
Converting back to Celsius: $T_1 = 400 - 273 = 127^{\circ}C$.

(A) The maximum theoretical efficiency of a heat engine is given by the Carnot efficiency formula: $\eta_{max} = 1 - \frac{T_L}{T_H}$.
Here,the source temperature $T_H = 127^{\circ}C = 127 + 273 = 400 \text{ K}$.
The sink temperature $T_L = 27^{\circ}C = 27 + 273 = 300 \text{ K}$.
Substituting these values: $\eta_{max} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$ or $25\%$.
Since the scientist claims an efficiency of $26\%$,which is greater than the maximum possible Carnot efficiency of $25\%$,this claim violates the second law of thermodynamics.
Therefore,this is impossible.

(A) The temperatures are given as $T_2 = 27 \, ^\circ C = 300 \, K$ and $T_1 = 927 \, ^\circ C = 1200 \, K$.
Efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{1200} = 1 - 0.25 = 0.75 = \frac{3}{4}$.
Also,efficiency $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat absorbed from the source.
Given $W = 12.6 \times 10^{6} \, J$.
Substituting the values: $\frac{3}{4} = \frac{12.6 \times 10^{6}}{Q_1}$.
Therefore,$Q_1 = \frac{4 \times 12.6 \times 10^{6}}{3} = 4 \times 4.2 \times 10^{6} = 16.8 \times 10^{6} \, J$.

(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 source temperature and $T_2$ is the sink temperature in Kelvin.
First,convert the temperatures from Celsius to Kelvin:
$T_1 = 727 + 273 = 1000 \ K$
$T_2 = 227 + 273 = 500 \ K$
Now,substitute these values into the efficiency formula:
$\eta = 1 - \frac{500}{1000} = 1 - 0.5 = 0.5$
Therefore,the maximum possible efficiency is $1/2$.

(C) The efficiency of a Carnot engine is given by $\eta = (1 - \frac{T_L}{T_H}) \times 100$,where $T_L$ is the sink temperature and $T_H$ is the source temperature.
For the first case: $40 = (1 - \frac{T_L}{500}) \times 100$.
$0.4 = 1 - \frac{T_L}{500} \implies \frac{T_L}{500} = 0.6 \implies T_L = 300 \ K$.
For the second case,the sink temperature $T_L$ remains constant at $300 \ K$ and efficiency $\eta$ becomes $60\%$:
$60 = (1 - \frac{300}{T_H'}) \times 100$.
$0.6 = 1 - \frac{300}{T_H'} \implies \frac{300}{T_H'} = 0.4$.
$T_H' = \frac{300}{0.4} = 750 \ K$.

(C) For a diatomic gas,the adiabatic index $\gamma = 1.4 = 7/5$.
In a Carnot cycle,the adiabatic expansion occurs between temperatures $T_1$ (source) and $T_2$ (sink).
The relation for adiabatic process is $T V^{\gamma - 1} = \text{constant}$.
Thus,$T_1 V_B^{\gamma - 1} = T_2 V_C^{\gamma - 1}$,where $V_B = V$ and $V_C = 32V$.
$\frac{T_2}{T_1} = \left( \frac{V_B}{V_C} \right)^{\gamma - 1} = \left( \frac{V}{32V} \right)^{1.4 - 1} = \left( \frac{1}{32} \right)^{0.4} = \left( \frac{1}{32} \right)^{2/5}$.
Since $32 = 2^5$,we have $\frac{T_2}{T_1} = (2^{-5})^{2/5} = 2^{-2} = 1/4 = 0.25$.
The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.
$\eta = 1 - 0.25 = 0.75$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the sink temperature and $T_1$ is the source temperature.
Given $T_2 = 7^{\circ}C = 280 \ K$ and $\eta_1 = 0.5$.
$0.5 = 1 - \frac{280}{T_1} \Rightarrow \frac{280}{T_1} = 0.5 \Rightarrow T_1 = 560 \ K$.
Now,for $\eta_2 = 0.7$ with the same sink temperature $T_2 = 280 \ K$:
$0.7 = 1 - \frac{280}{T_1'} \Rightarrow \frac{280}{T_1'} = 0.3 \Rightarrow T_1' = \frac{280}{0.3} = 933.33 \ K$.
The increase in source temperature is $\Delta T = T_1' - T_1 = 933.33 - 560 = 373.33 \ K$.
Rounding to the nearest given option,the increase is approximately $380 \ K$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q}$.
Initially,$\eta = 1/6$,so $1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_1 = \frac{6}{5}T_2$.
When the sink temperature is reduced by $62^{\circ}C$,the new sink temperature is $T_2' = T_2 - 62$.
The new efficiency is $\eta' = 2\eta = 2(1/6) = 1/3$.
Thus,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3}$.
Substituting $T_1 = \frac{6}{5}T_2$ into the equation: $\frac{T_2 - 62}{(6/5)T_2} = \frac{2}{3}$.
$\frac{5(T_2 - 62)}{6T_2} = \frac{2}{3} \implies 15(T_2 - 62) = 12T_2 \implies 15T_2 - 930 = 12T_2$.
$3T_2 = 930 \implies T_2 = 310 \text{ K}$.
Converting to Celsius: $T_2 = 310 - 273 = 37^{\circ}C$.
Now,$T_1 = \frac{6}{5} \times 310 = 6 \times 62 = 372 \text{ K}$.
Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ}C$.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.
Initially,$\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6} \quad \dots(i)$
When the sink temperature $T_2$ is reduced by $62^{\circ}C$ (which is equivalent to $62 K$ in absolute scale),the new sink temperature is $T_2' = T_2 - 62$.
The new efficiency is $\eta' = 1 - \frac{T_2 - 62}{T_1} = 2\eta$.
Substituting $\eta = 1/6$,we get $\eta' = 2 \times (1/6) = 1/3$.
So,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3} \quad \dots(ii)$.
From $(i)$,$\frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$.
Substitute $T_2$ into $(ii)$: $\frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{2}{3} \implies \frac{5}{6} - \frac{62}{T_1} = \frac{2}{3}$.
$\frac{62}{T_1} = \frac{5}{6} - \frac{4}{6} = \frac{1}{6} \implies T_1 = 62 \times 6 = 372 K$.
Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ}C$.
Now,$T_2 = \frac{5}{6} \times 372 = 310 K$.
Converting to Celsius: $T_2 = 310 - 273 = 37^{\circ}C$.
Thus,the sink temperature is $37^{\circ}C$ and the source temperature is $99^{\circ}C$.

(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 $\eta = 0.40$ and $T_2 = 300 \ K$,we have $0.40 = 1 - \frac{300}{T_1}$.
$\frac{300}{T_1} = 0.60 \implies T_1 = \frac{300}{0.60} = 500 \ K$.
Now,the efficiency is increased by $50\%$ of its original value. The new efficiency $\eta' = \eta + 0.50 \times \eta = 0.40 + 0.20 = 0.60$.
Keeping $T_2 = 300 \ K$ constant,the new source temperature $T_1'$ is given by $\eta' = 1 - \frac{T_2}{T_1'}$.
$0.60 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.40$.
$T_1' = \frac{300}{0.40} = 750 \ K$.
The increase in source temperature is $\Delta T = T_1' - T_1 = 750 \ K - 500 \ K = 250 \ K$.

(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.
Given $\eta = 0.5$ and $T_2 = 500 \ K$,we have $0.5 = 1 - \frac{500}{T_1}$,which gives $\frac{500}{T_1} = 0.5$,so $T_1 = 1000 \ K$.
Now,keeping the source temperature $T_1 = 1000 \ K$ constant,we want to achieve an efficiency $\eta' = 0.6$ by changing the sink temperature to $T_2'$.
Using the formula $\eta' = 1 - \frac{T_2'}{T_1}$,we get $0.6 = 1 - \frac{T_2'}{1000}$.
Rearranging the terms,$\frac{T_2'}{1000} = 1 - 0.6 = 0.4$.
Therefore,$T_2' = 0.4 \times 1000 = 400 \ K$.

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the intake temperature and $T_2$ is the exhaust temperature.
For the first case: $\eta_1 = 0.4$,$T_1 = 500 \ K$.
$0.4 = 1 - \frac{T_2}{500} \implies \frac{T_2}{500} = 0.6 \implies T_2 = 300 \ K$.
Since the exhaust temperature $T_2$ remains the same for the second case:
For the second case: $\eta_2 = 0.5$,$T_2 = 300 \ K$.
$0.5 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.5 \implies T_1' = \frac{300}{0.5} = 600 \ K$.
Thus,the new intake temperature is $600 \ K$.

(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L = 300 \ K$ and $T_H = 600 \ K$.
$\eta = 1 - \frac{300}{600} = 1 - 0.5 = 0.5$.
Efficiency is also defined as the ratio of work done $W$ to the heat supplied $Q_H$,i.e.,$\eta = \frac{W}{Q_H}$.
Given $W = 800 \ J$,we have $0.5 = \frac{800}{Q_H}$.
Therefore,$Q_H = \frac{800}{0.5} = 1600 \ J/cycle$.

(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 in Kelvin.
Given: $T_1 = 627^{\circ}C = 627 + 273 = 900 \ K$ and $T_2 = 227^{\circ}C = 227 + 273 = 500 \ K$.
Substituting the values: $\eta = 1 - \frac{500}{900} = 1 - \frac{5}{9} = \frac{4}{9}$.
To express as a percentage: $\eta = \frac{4}{9} \times 100 \% \approx 44.44 \%$.
Therefore, the correct option is $C$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the sink temperature and $T_1$ is the source temperature.
Given $\eta = 40\% = 0.4$ and $T_2 = 300\, K$.
$0.4 = 1 - \frac{300}{T_1} \implies \frac{300}{T_1} = 0.6 \implies T_1 = \frac{300}{0.6} = 500\, K$.
The efficiency is to be increased by $50\%$ of its original efficiency.
New efficiency $\eta' = \eta + 0.5 \times \eta = 0.4 + 0.5(0.4) = 0.4 + 0.2 = 0.6 = 60\%$.
Let the new source temperature be $T_1'$.
$0.6 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.4 \implies T_1' = \frac{300}{0.4} = 750\, K$.
The increase in source temperature is $\Delta T = T_1' - T_1 = 750 - 500 = 250\, K$.

(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 in Kelvin.
Given $\eta_1 = 1/6$,so $1/6 = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = 5/6$ or $T_2 = \frac{5}{6}T_1$ $...(i)$.
When the sink temperature is reduced by $62^{\circ}C$ (which is equivalent to a change of $62 \ K$),the new efficiency $\eta_2 = 2 \times \eta_1 = 2 \times (1/6) = 1/3$.
The new sink temperature is $T_2' = T_2 - 62$.
Using the efficiency formula: $1/3 = 1 - \frac{T_2 - 62}{T_1}$.
Rearranging gives $\frac{T_2 - 62}{T_1} = 1 - 1/3 = 2/3$.
Substituting $T_2 = \frac{5}{6}T_1$ from equation $(i)$:
$\frac{\frac{5}{6}T_1 - 62}{T_1} = 2/3$.
$\frac{5}{6} - \frac{62}{T_1} = 2/3$.
$\frac{62}{T_1} = \frac{5}{6} - \frac{4}{6} = 1/6$.
$T_1 = 62 \times 6 = 372 \ K$.
To convert to Celsius: $T(^{\circ}C) = 372 - 273 = 99^{\circ}C$.