A) 372K
B) 272K
C) 172K
D) 62K
Correct Answer: A
Solution :
Let the temperature of source and sink be \[{{T}_{1}}\] and \[{{T}_{2}}\](K) Efficiency in first case: \[{{\eta }_{1}}=\frac{1}{6}=\frac{({{T}_{1}}-{{T}_{2}})}{{{T}_{1}}}\] Efficiency in second case: \[{{\eta }_{2}}=2\times \frac{1}{6}=\frac{{{T}_{1}}-({{T}_{2}}-62)}{{{T}_{1}}}\] [Temperature change in \[{}^\circ C\] and K will be same] or \[\frac{1}{3}=\frac{({{T}_{1}}-{{T}_{2}})}{{{T}_{1}}}+\frac{62}{{{T}_{1}}}\] \[\frac{62}{{{T}_{1}}}\,=\frac{1}{3}\,-\frac{1}{6}\,=\frac{1}{6}\] \[\therefore \,\,\,\,\,\,\,\,{{T}_{1}}\,=62\times 6\,=372K\]
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