Question

The efficiency of Carnot’s heat engine is 0.5 when the temperature of the source is T1 and that of sink is T2. The efficiency of another Carnot’s heat engine is also 0.5. The temperature of source and sink of the second engine are respectively

• A. 2T1, 2T2
• B. 2T1, $\frac{T_{2}}{2}$
• C. T1 + 5, T2 – 5
• D. T1 + 10, T2 – 10

Answer 1

Answer: (A)

2T1, 2T2

Answer 2

Efficiency of carnot engine

$\eta = 1 - \frac{T_{2}}{T_{1}}$

For first Carnot’s heat engine, $0.5 = 1 - \frac{T_{2}}{T_{1}}$

For second heat engine, $0.5 = 1 - \frac{T'_{2}}{T'_{1}}$

Efficiency remains the same when both $T_{1}$and $T_{2}$ are increased by same factor

Therefore, the temperature of source and sink of second engine are

$T_{1}^{'} = 2T_{1},T_{2}^{'} = 2T_{2}$