Correct option: (D) 500 K and 300 K
Explanation:
Initial efficiency = η = 0.4
When T2 is lowered by 50k, η' = 0.5
η = 1 – (T2 / T1)……. efficiency of cannot engine
Hence 0.4 = 1 – (T2/T1) (1)
0.5 = 1 – [(T2 – 50) / T1] (2)
From (2)
0.5 = 1 – (T2 / T1) + (50 / T1)
From (1),
0.5 = 0.4 + (50 / T1)
∴ 0.1 = (50 / T1)
∴ T1 = 500K
From (1),
1 – 0.4 = [T2 / (500)] hence T2 = 500 × 0.6 = 300K