First slide

Heat engine

Question

A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of the source be increased so that its efficiency is increased 50% of original efficiency?

Moderate

A

275 K
B

325 K
C

250 K
D

380 K

Solution

The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied, i.e.,
$η = \frac{Work done}{Heat supplied} = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}}$
= 1- $\frac{Q_{2}}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}$
Here, $T_{1}$ is the temperature of source and $T_{2}$ is the temperature of the sink
As given, $η = 40 % = \frac{40}{100} = 0.4$
and $T_{2} = 300 K$
So, $0.4 = 1 - \frac{300}{T_{1}}$
$T_{1} = \frac{300}{1 - 0.4} = \frac{300}{0.6} = 500 K$
Let temperature of the source be increased by x K, then efficiency becomes
$η^{'} = 40 % + 50 % of η$
= $\frac{40}{100} + \frac{50}{100} \times 0.4 = 0.4 + 0.5 \times 0.4 = 0.6$
Hence, $0.6 = 1 - \frac{300}{500 + x}$
$\frac{300}{500 + x} = 0.4$
$500 + x = \frac{300}{0.4} = 750$
x = 750 - 500 = 250 K

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