Question

A Carnot's engine has an efficiency of $50\%$ at sink temperature $ 50{}^\circ C $ . Calculate the temperature of source.

• A. $ 133{}^\circ C $
• B. $ 143{}^\circ C $
• C. $ 100{}^\circ C $
• D. $ 373{}^\circ C $

Answer 1

Answer: (D)

$\eta=\frac{\text { Net work done per cycle }}{\text { Total amount of heat absorbed per cycle }}$
or $ \eta =\frac{W}{{{Q}_{1}}} $
$ \therefore $ $ \eta =\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}} $
Or $ \eta =1-\frac{{{Q}_{2}}}{{{Q}_{1}}} $
as $ \frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}} $
$ \therefore $ $ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}} $ ..(i)
where, $ {{T}_{2}} $ is temperature of sink and $ {{T}_{1}} $ is the temperature of the sources.
Here, $ \eta =50 $%
$ {{T}_{2}}=50{}^\circ C=273+50 $
$ =323K $
Putting these values in Eq. (i), we get
$ \frac{50}{100}=1-\frac{323}{{{T}_{1}}} $
$ T=646K=373{}^\circ C $