# 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?

arun
123 Points
6 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