Question

The efficiency of a heat engine working between the freezing point and boiling point of water is-
(A). $6.25\mathrm{\%}$
(B). $26.8\mathrm{\%}$
(C). $12.5\mathrm{\%}$
(D). $20\mathrm{\%}$

Answer 1

Hint: A heat engine has three main parts; source, working substance and the sink. The efficiency of a heat engine tells us about its performance. It depends on the temperature of the source and the sink. Substituting the values for source and sink given in the problem, we can calculate the efficiency.
Formulas used:
$\eta =1-{\displaystyle \frac{T_2}{T_1}}$

Complete step by step answer:
A heat engine consists of three parts-
Source- it is the part of the engine which is at higher temperature, let its temperature be $T_1$.
Working substance- it is the part of the engine through which heat changes take place. It is usually an ideal gas.
Sink- it is the part of the heat engine which is at lower temperature, let its temperature be $T_2$.
The efficiency of a heat engine tells us how well a heat engine performs; the efficiency depends on the temperature of the source and the sink, therefore,
$\eta =1-{\displaystyle \frac{T_2}{T_1}}$
Here, $\eta$ is the efficiency of the heat engine
Given, the freezing point of water is at $273K$, hence $T_2=273K$, the boiling point of water is at $373K$, hence $T_1=373K$
Substituting given values in the above equation, we get,
$\eta =1-{\displaystyle \frac{273}{373}}$
$\Rightarrow \eta ={\displaystyle \frac{100}{373}}\times 100\mathrm{\%}$
$\therefore \eta =26.8\mathrm{\%}$
So, the correct answer is “Option B”.

Note: In a heat engine, the working substance absorbs heat from the source and rejects heat to the sink while in a refrigerator; the working substance absorbs heat from the sink and rejects heat to the source. To analyse the performance of a refrigerator, coefficient of performance is calculated. The efficiency of a heat engine is a unitless quantity.