A) \[62{}^\circ C\]
B) \[99{}^\circ C\]
C) \[37{}^\circ C\]
D) \[124{}^\circ C\]
Correct Answer: B
Solution :
Efficiency of an engine \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[{{T}_{1}}\,\,=\, temperature of source\] \[{{T}_{2}} = temperature of sink\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{6}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{5}{6}\] When temperature of sink is decreased by \[62{}^\circ C\](or 62 K), efficiency becomes double, as temperature of source remain un change \[\therefore \,\,\,\,\,\,\,\,\,\,2\times \frac{1}{6}=\frac{1-({{T}_{1}}-62)}{{{T}_{1}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{1}{3}=1-\frac{({{T}_{2}}-62)}{{{T}_{1}}}\] ? (ii) Solving Eqs. (i) and (ii) We have \[{{\operatorname{T}}_{1}}=372 K\,\,= \,99{}^\circ C\]
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