Question

Two Carnot engines A and B are operated in series. The first one A receives heat from a reservoir at 800 K and rejects heat to a reservoir heat to reservoir at temperature T. The second Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 300 K. Calculate the temperature T for the following situations:
(A) The output of engines are equal.
(B) The efficiencies of two engines are equal.

Answer 1

Hint: When we will solve the first case when the outputs of two engines are equal. We have to keep in mind that the heat rejected by the first reservoir is taken as input by the second B reservoir and simply apply the formula. Now for the second case we have to simply apply the formula of efficiency of engines and equate it.

Formula used:
$W={{Q}_{1}}-{{Q}_{2}}$
$\dfrac{{{Q}_{2}}}{{{Q}_{1}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$

Efficiency of engine
$\eta =1-\dfrac{{{Q}_{2}}}{{{Q}_{1}}}=1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$


Complete answer:
 Let the output of both the engines be W.

Let the engine A take ${{Q}_{1}}$ heat as input at temperature ${{T}_{1}}$ and gives out heat ${{Q}_{2}}$ at temperature T the second engine B receive ${{Q}_{2}}$ at temperature ${{T}_{2}}$ to the sink.

Work done by engine, A $W={{Q}_{1}}-{{Q}_{2}}$
Work done by engine, B $W={{Q}_{2}}-{{Q}_{3}}$

Thus,
${{Q}_{1}}-{{Q}_{2}}={{Q}_{2}}-{{Q}_{3}}$

Dividing both the sides by${{Q}_{1}}$ we get,
$\begin{align}
  & 1-\dfrac{{{Q}_{2}}}{{{Q}_{1}}}=\dfrac{{{Q}_{2}}}{{{Q}_{1}}}-\dfrac{{{Q}_{3}}}{{{Q}_{1}}} \\
 & =1-\dfrac{T}{{{T}_{1}}}=\dfrac{{{Q}_{2}}}{{{Q}_{1}}}\left( 1-\dfrac{{{Q}_{3}}}{{{Q}_{2}}} \right) \\
 & =1-\dfrac{T}{{{T}_{1}}}=\dfrac{{{Q}_{2}}}{{{Q}_{1}}}\left( 1-\dfrac{{{T}_{3}}}{T} \right) \\
 & =1-\dfrac{T}{{{T}_{1}}}=\dfrac{T}{{{T}_{1}}}\left( 1-\dfrac{{{T}_{3}}}{T} \right) \\
 & =\dfrac{{{T}_{1}}}{T}-1=1-\dfrac{{{T}_{3}}}{T} \\
 & =\dfrac{{{T}_{1}}}{T}+\dfrac{{{T}_{3}}}{T}=2 \\
 & =\dfrac{1}{T}\left( {{T}_{1}}+{{T}_{3}} \right)=2 \\
 & =T=\dfrac{\left( {{T}_{1}}+{{T}_{3}} \right)}{2} \\
 & =T=\dfrac{\left( 800+300 \right)}{2}=550K \\
\end{align}$

550K is the temperature when the output of engines are equal.

(B)
Let the efficiency of both engines be $\eta $.
Now considering both engines efficiency are equal. This gives

$\begin{align}
  & 1-\dfrac{T}{{{T}_{1}}}=1-\dfrac{{{T}_{3}}}{T} \\
 & =\dfrac{T}{{{T}_{1}}}=\dfrac{{{T}_{3}}}{T} \\
 & ={{T}^{2}}={{T}_{1}}\times {{T}_{3}} \\
 & ={{T}^{2}}=800\times 300=240000\\
\end{align}$

T is 489.897K
Temperature when both the engines have equal efficiencies is 489.897K

Additional Information:
The efficiency of a carnot engine:
Depends upon the temperatures of the source and the sink. It is independent of the nature of the working substance. It is the same for all reversible engines working between the same temperatures. Is directly proportional to the temperature difference.
Carnot engine is an ideal reversible heat engine that operates between two temperatures ${{T}_{1}}$ source and ${{T}_{2}}$ sink.

Note:
As
 $\begin{align}
  & \dfrac{{{Q}_{2}}}{{{Q}_{1}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}} \\
 & \therefore \text{if }{{T}_{2}}=0K\text{ ,Then }{{\text{Q}}_{2}}=0 \\
\end{align}$
Since ${{T}_{2}}=0K$ cannot be realized, so ${{\text{Q}}_{2}}=0$ is also not possible. This means that it is not possible to convert the whole of heat energy absorbed from the source into mechanical work continuously, without rejecting a part of it to the sink.
If ${{Q}_{1}}$ and ${{Q}_{2}}$ are in joule and temperature ${{T}_{1}}$ and ${{T}_{2}}$ are in kelvin. Efficiency $\eta $ has no units.