A) 225 K
B) \[128{}^\circ C\]
C) 580 K
D) 145 K
E) \[145{}^\circ C\]
Correct Answer: D
Solution :
\[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] Initially, \[\frac{50}{100}=1-\frac{(273+17)}{{{T}_{1}}}\] Or \[\frac{290}{{{T}_{1}}}=\frac{1}{2}\] \[\Rightarrow \] \[{{T}_{1}}=580\,K\] Finally, \[\frac{60}{100}=1-\frac{(273+17)}{T_{1}^{}}\] Or \[\frac{290}{T_{1}^{}}=\frac{2}{5}\] \[\Rightarrow \] \[T_{1}^{}=725\,K\] Hence, change in source temperature \[=(725-580)K=145K\]
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