Question

A Carnot engine absorbs 300cal of heat from a reservoir and rejects 250cal of heat at 300K during each cycle. calculate:. 1 the efficiency of engine 2 the Temperature of reservior​

Answer 1

Q2 = 250 Cal     T2 = 300°K
Work done = 300 - 250 = 50 Cal

Efficiency = η = W/Q1 = 1 -  Q2/Q1  = 1 - 250/300 = 1/6
                = 16.67 %

Q2/Q1 = T2/T1
=>   T1 = 300 K * 300 cal /250cal
           = 360°K