First slide

Heat engine

Question

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

Easy

A

26.8 %
B

20 %
C

6.25 %
D

12.5 %

Solution

Efficiency of an ideal heat engine,
$η = (1 - \frac{T_{2}}{T_{1}})$
$Freezing point of water = 0 ° C = 273 K$
$Boiling point of water = 100 ° C = (100 + 273) K = 373 K$
$\begin{array}{l} T_{2} : Sink temperature = 273 K \\ T_{1} : Source temperature = 373 K \end{array}$
$% η = (1 - \frac{T_{2}}{T_{1}}) \times 100 = (1 - \frac{273}{373}) \times 100$
$= (\frac{100}{373}) \times 100 = 26.8 %$

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