A) 600 K
B) 800 K
C) 900 K
D) 1000 K
Correct Answer: A
Solution :
The efficiency of a Carnot engine operatic 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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