Question

The source and sink temperature of a Carnot engine are $400K$ and $300K$ respectively. What is the efficiency $?$
A) $100\%$
B) $75\%$
C) $33.3\%$
D) $25\%$

Answer 1

We know the temperature value of the hot and cold sink in this above statement, then we use the general efficiency formula to find the percentage of Carnot engine temperature efficiency. A theoretical engine that operates on a reversible Carnot cycle is a Carnot heat engine.
Useful formula:
The Carnot efficiency
$\eta = 1 - \dfrac{{{t_L}}}{{{T_H}}}$
Where,
${t_L}$ is the temperature of cold sink
${T_H}$ is temperature of hot sink
$\eta$ is efficiency
Complete step by step solution:
Given by,
${T_H} = 400K$
${t_L} = 300K$
These are the temperature of source and sink respectively
A Carnot cycle is an ideal thermodynamic reversible closed cycle in which four successive operations are involved, including isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
The efficiency of the cycle is,
$\eta = 1 - \dfrac{{{t_L}}}{{{T_H}}}$
During these operations, the expansion and compression of substance can be done up to the desired point and back to the initial state.
Now,
Substituting the given value in above equation
$\eta = 1 - \dfrac{{300}}{{400}}$
On simplifying,
We get,
$\eta = 1 - 0.75$
The above equation can be solved,
$\eta = 0.25$
Thus, the efficiency is $25\%$
Hence, the option D is correct answer

Note: when we have to measure the efficiency, we know the temperature value then we provide the estimation of the possible efficiency that a heat engine will possess during the heat-to-work conversion process and, conversely, operating between two reservoirs. The performance of this type of engine is also independent of the existence of the working substance and depends only on the hot and cold reservoir temperature.