A reversible heat engine operates with an efficiency of 50%. If during each cycle it rejects 150 cal to a reservoir of heat at 30°C, then what is the temperature of the other reservoir and how much work does it carry out per cycle?

Answer 1

.5=(tin-tout)/tin
where tout is (30+273) , solve for tin (temp of original reservoir)

work per cycle=300cal (to get 50 percent efficiency)