A) 840 K
B) 280 K
C) 560 K
D) 380 K
Correct Answer: D
Solution :
\[{{\operatorname{T}}_{2}}=7{}^\circ C=(7+273)=280K\] \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\eta =1-\frac{50}{100}=\frac{50}{100}=\frac{1}{2}\] \[\therefore \,\,\,\,{{T}_{1}}=2\times {{T}_{2}}=2\times 280=560\,K\] New efficiency, \[\eta =70\,%\] \[\therefore \,\,\,\,{{T}_{1}}=\frac{10}{3}\times 280=\frac{2800}{3}=933.3\,K\] \[\therefore \] Increase in the temperature of high temp. \[\operatorname{reservoir}=933.3-560=373.3K=380K\]
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