A) \[{{90}^{\text{o}}}C,\,{{37}^{\text{o}}}C\]
B) \[{{99}^{\text{o}}}C,\,{{37}^{\text{o}}}C\]
C) \[{{372}^{\text{o}}}C,\,{{37}^{\text{o}}}C\]
D) \[{{206}^{\text{o}}}C,\,{{37}^{\text{o}}}C\]
Correct Answer: B
Solution :
Key Idea: The efficiency of heat engine is the ratio of work done to the heat taken from the source. If \[{{T}_{1}}\] is temperature of source and \[{{T}_{2}}\] the temperature of sink, the efficiency of engine \[\eta =\frac{Work\,done(W)}{Heat\,taken\,({{Q}_{1}})}\] \[=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\therefore 1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{1}{6}\] ....(i) When temperature of sink is reduced by 62°C then \[T_{2}^{'}={{T}_{2}}-62\] \[\therefore \eta '=1-\frac{T_{2}^{'}}{{{T}_{1}}}\] \[Given\,:\,\eta '=2\eta =\frac{2}{6}=\frac{1}{3}\] \[\therefore \frac{1}{3}=1-\frac{{{T}_{2}}-62}{{{T}_{1}}}...(ii)\] From Eq. (i) \[\frac{{{T}_{2}}-62}{{{T}_{1}}}=\frac{2}{3}\] .....(iv) Dividing Eq. (iii) by Eq. (iv) \[\frac{{{T}_{2}}}{{{T}_{2}}-62}=\frac{5}{4}\] \[\Rightarrow 4{{T}_{2}}=5{{T}_{2}}-310\] \[\Rightarrow {{T}_{2}}=310\,K\] and from Eq. (iii), we have \[\frac{310}{{{T}_{1}}}=\frac{5}{6}\] \[\Rightarrow {{T}_{1}}=372\,K\] Hence, \[{{T}_{1}}=372\,K=372-273={{99}^{o}}C\] and \[{{T}_{2}}=310\,K=310-273={{37}^{o}}C\]
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