The temperature of the source of a carnot engine is `T_1` and that of it's sink is `T_2`. It's efficiency is 0.5. The temperature of source and sink of another carnot engine of same efficiency as the first one, are respectively:

Options

A. `2T_1, 2T_2`
B. `2T_1, (T_2)/2`
C. `T_1 + 5, T_2 - 5`
D. `T_1 + 10, T_2 - 10`

Answer

A

Explanation

Efficiency of Carnot.s heat engine,
`eta = 1 - (T_2)/(T_1)`
for first Carnot.s heat engine,
`0.5 =1 - (T_2)/(T_1)`
For second heat engine, `0.5 = 1 - (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 áre `T_1. = 2T_1, T_2 = 2T_2`