Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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 temperatures of the source and the sink will be
(A) $375K$ , $300K$
(B) $475K$ , $400K$
(C) $400K$ , $800K$
(D) $375K$ , $400K$

seo-qna
Last updated date: 17th Apr 2024
Total views: 35.7k
Views today: 0.35k
Answer
VerifiedVerified
35.7k+ views
Hint: - The efficiency of Carnot’s heat engine depends on the temperature of the heat source and that of the heat sink. The greater the temperature between the source and the sink, the 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:
A 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 founded 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
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ .......... $\left( 1 \right)$
where ${T_1}$ is the temperature of the source and ${T_2}$ is the temperature of the sink.
It is given that the efficiency between the two temperatures is $0.2$ . Therefore, by equation $\left( 1 \right)$
$0.2 = 1 - \dfrac{{{T_2}}}{{{T_1}}} \to {T_2} = 0.8{T_1}$ ............ $\left( 2 \right)$
Also, when the temperature of the source ${T_1}$ is increased by ${25^ \circ }C$, the efficiency increases to $0.25$. Thus, the increase in efficiency can be written as
$0.25 = 1 - \dfrac{{{T_2}}}{{{T_1} + 25}}$
$ \Rightarrow \dfrac{{{T_2}}}{{{T_1} + 25}} = 1 - 0.25 = 0.75$
Substituting the value of ${T_2}$ from equation $\left( 2 \right)$
$\dfrac{{0.8{T_1}}}{{{T_1} + 25}} = 0.75$
$ \Rightarrow 0.8{T_1} = 0.75\left( {{T_1} + 25} \right)$
Solving the equation for ${T_1}$ we get,
$0.8{T_1} = 0.75{T_1} + \left( {0.75 \times 25} \right)$
$ \Rightarrow 0.05{T_1} = 18.75$
On further solving the equation we get,
${T_1} = \dfrac{{18.75}}{{0.05}}$
$ \Rightarrow {T_1} = 375K$
The value of ${T_2}$ can be derived by equation $\left( 2 \right)$
${T_2} = 0.8{T_1} = 0.8 \times 375 = 300K$
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. That’s why it is said to be an ideal heat engine.