Question

The efficiency of a Carnot engine world between two temperatures is 0.2, when the temperature of the source is increased by ${25^ \circ }C$, the efficiency increases to 0.25. The
temperature of the source and sink will be
A. 375K, 300K
B. 475K, 400K
C. 400K, 800K
D. 375K, 400K

Answer 1

The efficiency of Carnot’s heat engine depends on the temperature of the heat source
and that of the heat sink. Greater the temperature between source and the sink, higher will be the
efficiency. If the temperature of the source is increased by${25^ \circ }C$ , then the efficiency will also increase.

Formula Used: The efficiency of Carnot’s heat engine is: $\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$

Complete step by step solution: Heat engine is a system or a device operating in a cyclic process and converting heat into work, without itself undergoing any change at the end of the cycle. It is based on the principle that a system whose different parts are at different temperatures tends to change towards a thermodynamic equilibrium state.

The Carnot heat engine is an ideal heat engine. The efficiency of Carnot’s heat engine is
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ $\to (1)$
where ${T_1}$ is the temperature of source and ${T_2}$ is the temperature of sink.
It is given that the efficiency between two temperatures is 0.2. Therefore, by equation (1)
$0.2 = 1 - \dfrac{{{T_2}}}{{{T_1}}} \to \dfrac{{{T_2}}}{{{T_1}}} = 0.8 \to {T_2} = 0.8{T_1}$ $\to (2)$
Also, when the temperature of the source ${T_1}$ is increased by ${25^ \circ }C$, the efficiency
increases to 0.25. Therefore, the increase in efficiency can be written as
$0.25 = 1 - \dfrac{{{T_2}}}{{{T_1} + 25}}$
$\to \dfrac{{{T_2}}}{{{T_1} + 25}} = 1 - 0.25 = 0.75$
Substituting the value of ${T_2}$ from equation (2)
$\to \dfrac{{0.8{T_1}}}{{{T_1} + 25}} = 0.75$

\to 0.8{T_1} = 0.75\left( {{T_1} + 25} \right) \\\ \to 0.8{T_1} = 0.75{T_1} + \left( {0.75 \times 25} \right) \\\ \to 0.05{T_1} = 18.75 \\\ \to {T_1} = \dfrac{{18.75}}{{0.05}} \\\ \to {T_1} = 375 \\\ \end{gathered} $$ The value of $${T_2}$$ can be derived by equation (2) $${T_2} = 0.8{T_1} = 0.8 \times 375 = 300$$ Therefore, the temperature of the source and sink will be $$375K$$ and $$300K$$. Hence, option (A) is the correct answer. **Note:** The requirements for the Carnot heat engine are difficult rather impossible to be satisfied in actual practice. Therefore the Carnot heat engine can be designed only in thought. For this reason, it is said to be an ideal heat engine.