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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

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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