Question

An ideal refrigerator has a freezer at a temperature of – 13 C, The coefficient of performance of the engine is 5. The temperature of the air to which heat is rejected will be. 

(A) 325°C 

(B) 39°C 

(C) 325 K 

(D) 320°C

Answer 1

Correct option: (B) 39°C

Explanation:

Given : temperature of freezer = T₂­ = – 13 °C

= – 13 + 273 = 260K

Coefficient of performance = β = 5

And      β = [T₂ / (T₁ – T₂)]

∴  5 = [(260) / (T₁ – 260)]

∴   T₁ = 260 + 52 = 312K = (312 – 273) °C = 39°C