Let ₁ be the efficiency of a Carnot engine at T_H = 547^ C and T_C = 247^ C while ₂ is the efficiency at T_H = 847^ C and T_C = 147^ C . If ₁ = x 41 ₂ , the value of x is

Step-by-step solution

The temperatures for the first Carnot engine in Kelvin are: T_ H1 = 547^ C + 273 = 820 K T_ C1 = 247^ C + 273 = 520 K The efficiency ₁ is given by: ₁ = 1 - T_ C1 T_ H1 = 1 - 520 820 = 1 - 26 41 = 15 41 The temperatures for the second Carnot engine in Kelvin are: T_ H2 = 847^ C + 273 = 1120 K T_ C2 = 147^ C + 273 = 420 K The efficiency ₂ is given by: ₂ = 1 - T_ C2 T_ H2 = 1 - 420 1120 = 1 - 3 8 = 5 8 Given the relation ₁ = x 41 ₂ , substituting the values of ₁ and ₂ : 15 41 = x 41 5 8 15 = 5x 8 x = 15 8 5 = 24 Answe