Question

For a Carnot engine, the source is at 500K and the sink at 300K. What is the efficiency of this engine?
A. 0.2
B. 0.4
C. 0.6
D. 0.3

Answer 1

Hint: Efficiency of a Carnot engine is the fraction of the heat absorbed by an engine that it can convert into work.
It is denoted by \[{\rm{\eta }}\].

Formula Used:
\[{\rm{\eta = }}\left( {\dfrac{{{{\rm{T}}_{\rm{1}}}{\rm{ - }}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}} \right)\]
where
\[{{\rm{T}}_{\rm{1}}}\]=temperature of the source
\[{{\rm{T}}_2}\]=temperature of the sink

Complete step-by-step answer:
The Carnot engine works on the principle of the second law of thermodynamics.
The efficiency of a Carnot engine denoted by η is the fraction of the heat absorbed by an engine transformed into work.
In the given question, we have
\[{{\rm{T}}_{\rm{1}}}\]=temperature of the source=500 K
\[{{\rm{T}}_2}\]=temperature of the sink=300 K
\[{\rm{\eta = }}\left( {\dfrac{{{{\rm{T}}_{\rm{1}}}{\rm{ - }}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}} \right)\]
\[{\rm{ = }}\left( {\dfrac{{{\rm{500K - 300K}}}}{{500{\rm{K}}}}} \right)\]
\[{\rm{ = }}\left( {\dfrac{{{\rm{200K}}}}{{500{\rm{K}}}}} \right)\]
=0.4

So, option B is correct.

Additional Information: (T₁-T₂T₁) is a fraction so it will always be less than 1.
It means the efficiency of a heat engine is always less than 1. No heat engine can have an efficiency equal to unity.
The efficiency depends on the difference between T₁ and T₂.
Thus, the greater the difference between the temperature of the source and the sink, the greater the efficiency.
This is why superheated steam is used in a steam engine.
All the periodic machines working reversibly between the same two temperatures have the same efficiency.

Note: While attempting the question, one must indicate the unit of temperature in the calculation step. Efficiency does not have any unit. So, the final answer will not have any unit.