A) \[90\,{}^\circ C,\text{ }37\,{}^\circ C\]
B) \[99\,{}^\circ C,\text{ }37\,{}^\circ C\]
C) \[372\,{}^\circ C,\text{ }37\,{}^\circ C\]
D) \[206\,{}^\circ C,\text{ }37\,{}^\circ C\]
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 |
\[\eta =\frac{\text{Work}\,\text{done}\,(W)}{\text{Heat}\,\text{taken}\,({{Q}_{1}})}\] |
\[=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] |
\[\therefore \] \[1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{1}{6}\] ....(i) |
When temperature of sink is reduced by \[62{{\,}^{o}}C\]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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