A) 600 K
B) 800 K
C) 900 K
D) 1000 K
Correct Answer: A
Solution :
The efficiency of a C are not engine operating between temperatures \[{{T}_{1}}\]and \[{{T}_{2}}\]is given by \[\eta -1-\frac{{{T}_{2}}}{{{T}_{1}}}\] where \[{{T}_{1}}=\]source temperature \[{{T}_{2}}=\]sink temperature In 1st case: \[\frac{40}{100}=1-\frac{{{T}_{2}}}{500}\] or \[\frac{{{T}_{2}}}{500}=1-0.4=0.6\] or \[{{T}_{2}}=0.6\times 500=300\,K\] In 2nd case: \[\frac{50}{100}\,=1-\frac{300}{T_{1}^{}}\] or \[\frac{300}{T_{1}^{}}\,=1-0.5\,=0.5\] or \[T_{1}^{}\,=\frac{300}{0.5}=600\,K\] Note: The efficiency of Carnot reversible engine is independent of the working substance and depends only on the absolute temperatures of the sink and the source.You need to login to perform this action.
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