Question

A Carnot engine, whose sink is at $300\,{\text{K}}$, has an efficiency of 40%. By how much the source temperature should be changed so as to increase the efficiency to 60%?
A. $250\,{\text{K}}$ increase
B. $250\,{\text{K}}$ decrease
C. $325\,{\text{K}}$ increase
D. $325\,{\text{K}}$ decrease

Answer 1

Use the formula for efficiency of the Carnot’s engine. This formula gives the relation between efficiency of Carnot’s engine, temperature of source and temperature of sink. First determine the temperature of source for first efficiency and then determine the change in temperature of the source that should be made.

Formula used:
The efficiency $\eta$ of the Carnot’s engine is given by
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ …… (1)
Here, ${T_1}$ is the temperature of the source and ${T_2}$ is the temperature of the sink.

Complete step by step answer:
We have given that the efficiency of the Carnot’s engine is $40\%$ when the temperature of the sink is $300\,{\text{K}}$.
$\eta = 40\% = \dfrac{{40}}{{100}} = 0.4$
${T_2} = 300\,{\text{K}}$

Let us first determine the temperature of source for this efficiency using equation (1).

Substitute $0.4$ for $\eta$ and $300\,{\text{K}}$ for ${T_2}$ in equation (1).
$0.4 = 1 - \dfrac{{300\,{\text{K}}}}{{{T_1}}}$
$\Rightarrow \dfrac{{300\,{\text{K}}}}{{{T_1}}} = 1 - 0.4$
$\Rightarrow {T_1} = \dfrac{{300\,{\text{K}}}}{{0.6}}$
$\Rightarrow {T_1} = 500\,{\text{K}}$

Hence, the temperature of source for efficiency 40% is $500\,{\text{K}}$.

We have given that the efficiency of the Carnot’s engine is increased to 60% for the same temperature of the sink.
$\eta ' = 60\% = \dfrac{{60}}{{100}} = 0.6$

For increasing this efficiency of the Carnot’s engine, let the temperature of the source is increased by $T$. Hence, the temperature of the source becomes ${T_1} + T$.
${{T_1}^1} = {T_1} + T$

Rewrite equation (1) for the new efficiency of the Carnot’s engine.
$\eta ' = 1 - \dfrac{{{T_2}}}{{T_1^1}}$

Substitute ${T_1} + T$ for $T_1^1$ in the above equation.
$\eta ' = 1 - \dfrac{{{T_2}}}{{{T_1} + T}}$

Substitute for $\eta '$, $300\,{\text{K}}$ for ${T_2}$ and $500\,{\text{K}}$ for ${T_1}$ in the above equation.
$0.6 = 1 - \dfrac{{300\,{\text{K}}}}{{\left( {500\,{\text{K}}} \right) + T}}$
$\Rightarrow \dfrac{{300\,{\text{K}}}}{{\left( {500\,{\text{K}}} \right) + T}} = 1 - 0.6$
$\Rightarrow 300\,{\text{K}} = 0.4\left( {\left( {500\,{\text{K}}} \right) + T} \right)$
$\Rightarrow 300 = 200 + 0.4T$
$\Rightarrow 0.4T = 300 - 200$
$\Rightarrow T = \dfrac{{100}}{{0.4}}$
$\therefore T = 250\,{\text{K}}$
Therefore, the temperature of the source should be increased by $250\,{\text{K}}$.

Hence, the correct option is A.

Note: The students may get confused that how we can determine the temperature of the source should be increased. We can see that the temperature of the sink is kept constant then also the efficiency of the engine is increased. In order to achieve this increase, the temperature of the source should be increased. If this efficiency was decreased at constant temperature of sink then the temperature of the source needs to be decreased.