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