A) \[\frac{100-23}{273}\]
B) \[\frac{100+23}{273}\]
C) \[\frac{100+23}{100}\]
D) \[\frac{100-23}{100}\]
Correct Answer: B
Solution :
Efficiency of Carnot engine is given by \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}\] Given, \[{{T}_{1}}=\]temperature of reservoir \[=100+273=373\,K\] \[{{T}_{2}}=\]temperature of sink \[=-\,23\,+\,273\] = 250 K Substituting in Eq.(i), we get \[\therefore \]\[\eta =\frac{373-250}{373}=\frac{123}{373}\] \[=\frac{100+23}{373}\] Hence, the correction option is [b].
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