A) 107 K
B) 500 K
C) 140 K
D) 200 K
Correct Answer: C
Solution :
Given, \[\eta =50%=\frac{1}{2};{{T}_{2}}=273+7=280\,K\] now, \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] or \[\frac{1}{2}=1-\frac{280}{{{T}_{1}}}\] \[\Rightarrow \] \[{{T}_{1}}=280\times 2=560\,K\] Now, \[\eta =50+10=60%=\frac{3}{5};{{T}_{2}}'=280K\] \[\therefore \] \[\eta =1-\frac{{{T}_{2}}'}{{{T}_{1}}'}\] \[\Rightarrow \] \[\frac{3}{5}=1-\frac{280}{{{T}_{1}}'}\] Or \[{{T}_{1}}'=\frac{5}{2}\times 280=700\,K\] \[\therefore \]Increase in temperature of hot reservoir, \[=700-560=140\text{ }K\]
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