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A reversible engines operates with an efficiency of 50% if during each cycle it rejects 150 cals to a reservoir of heat at 30C then,
1)what is the temperature of the other reservoir?
2)how much work does it carry out per cycle?

spursh , 9 Years ago
Grade 11
anser 1 Answers
arun

Last Activity: 9 Years ago

efficiency(\eta ) is 50% = 0.5
\eta = 1-\frac{T(sink)}{T(source)}
0.5 = 1-\frac{30}{T(source)}
0.5 = \frac{30}{T(source)}
T(source)=60 degree celcius
\eta = 1-\frac{Q(sink)}{Q(source)}
0.5 = 1-\frac{150}{Q(source)}
Q(source)=300 calories
work= Q(source) - Q(sink)
work= 150 calories
so temprature of other reservoir is 60 degree celcius
and work done by engine in one cycle is 150 calories

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