A Carnot engine has efficiency 1 6 . It becomes 1 3 , when the temperature of sink is lowered by 57 K . The temperature of the source is

Step-by-step solution

The Carnot efficiency is given by = 1 - T_L T_H where T_L and T_H are the sink and source temperatures in Kelvin. Given the initial efficiency ₁ = 1 6 , we have 1 6 = 1 - T_L T_H which simplifies to T_L T_H = 5 6 When the sink temperature decreases by 57 K and efficiency increases to ₂ = 1 3 , the relation becomes 1 3 = 1 - T_L - 57 T_H yielding T_L - 57 T_H = 2 3 Substituting T_L = 5 6 T_H from the first relation into the second: 5 6 T_H - 57 T_H = 2 3 Multiplying through by T_H : 5 6 T_H - 57 = 2 3 T_H Solving fo