A) 30%
B) 65%
C) 75%
D) 95%
Correct Answer: D
Solution :
Temperature of source \[{{T}_{1}}=6000K\] Temperature of sink \[{{T}_{2}}=300K\] Then efficiency \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{300}{6000}=\frac{19}{20}\] Efficiency percentage \[=\frac{19}{20}\times 100=95%\]You need to login to perform this action.
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