A Carnot engine converts one sixth of the heat input into work. If the sink temperature is reduced by 62^o C, the efficiency gets doubled. Find the source and the sink temperature.

Maximum efficiency of an engine working between temperatures T₂ and T₁ is given by the fraction of the heat absorbed by an engine which can be converted into work is known as efficiency of the heat engine.

Mathematically,
Efficiency, η = (T₂ – T₁) / T₂ = 1/6
6 T₂ -6 T₁ = T₂
Therefore, T₂ = 1.2 T₁ ………………….. (1)

Where T₂ is the source temperature and T₁ is the sink temperature.

If the sink temperature (T₁) is reduced by 62^o C, the efficiency gets doubled i.e.
η= T₂ – (T₁ – 62) / T₂ = 2 X 1/6 = 2/6
6 T₂ – 6 T₁ + 372 = 2 T₂
6 T₂ – 2 T₂ – 6 T₁ + 372 = 0
4 T₂ – 6 T₁ + 372 = 0

Substituting the value of T₂ from equation (1), we will get,
4 (1.2 T₁) – 6 T₁ + 372 = 0
4.8 T₁ – 6 T₁ + 372 = 0
372 = 1.2 T₁
T₁ = 310 K
Therefore, T₂ = 1.2 X 310 = 372 K
Hence, temperature of the source and the sink is 372 K and 310 K respectively.