A reversible Carnot heat engine converts 1 4 th of its input heat into work. When the temperature of the sink is reduced by 50 ~K , its efficiency becomes 33 1 3 % . The initial temperatures of the source and the sink respectively are

Step-by-step solution

Given work, W= Q 4 where, Q= input heat. Efficiency, = W Q = Q / 4 Q = 1 4 Also, =1- T₂ T₁ where, T₂= temperature of sink and T₁= temperature of source. 1- T₂ T₁ = 1 4 When temperature of sink is reduced by 50 ~K , efficiency becomes 100 3 % = 100 3 1 100 = 1 3 So, 1- T₂-50 T₁ = 1 3 Subtracting Eq (ii) from Eq. (i), we get aligned & 50 T₁ = 1 3 - 1 4 = 1 12 & T₁=600 ~K & T₂= 3 4 600=450 ~K aligned