Question

The maximum efficiency of a heat engine operating between ${100}^{{}^{\circ }}C$ and ${25}^{{}^{\circ }}C$
A. 20 %
B. 22.2%
C. 25%
D. None of the above

Answer 1

Hint: The Carnot cycle has the greatest efficiency possible of an engine because it is assumed that all other wasteful processes don't take place such as friction and no conduction of heat between different parts of the engine at different temperatures.

Complete Solution :
The efficiency can be calculated by the following formula:
Efficiency= ${\displaystyle \frac{T_2-T_1}{T_2}}$
Therefore, we have given temperatures ${100}^{{}^{\circ }}C$and ${25}^{{}^{\circ }}C$.
$T_1=25+273=298K$
$T_2=100+273=373K$
$\eta =$ ${\displaystyle \frac{T_2-T_1}{T_2}}$
$\Rightarrow {\displaystyle \frac{373-298}{373}}={\displaystyle \frac{75}{373}}$
$\Rightarrow 0.20\times 100$
$\Rightarrow 20$
Hence, 20% is the maximum efficiency of a heat engine operating between ${100}^{{}^{\circ }}C$ and ${25}^{{}^{\circ }}C$.
So, the correct answer is “Option B”.

Note: The Carnot cycle has four different process which leads to the maximum efficiency:
First process is a reversible isothermal gas expansion process. In this process, the ideal gas in the system absorbs q amount of heat from a heat source at a high temperature, expands and does work on surroundings.
- Reversible adiabatic gas expansion process, in this the system is thermally insulated and gas continues to expand and do work on the surrounding which causes the system to lower the temperature.
- Reversible isothermal gas compression: In this process, surroundings do the work to the gas and causes a loss of heat q.
- Reversible adiabatic gas compression: In this process, the system is thermally insulated. Surroundings continue to do the work to the gas which again causes the rise in the temperature. These all steps complete a Carnot engine with efficiency.