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.
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