Question

A cannot heat engine whose sink is at 200 K, has an efficiency 30%. By how much the temperature of the source be increased to have its efficiency equal to 50%. Keeping sink temperature constant.

Answer 1

From the relation,

η = 1 − \(\frac{T_2}{T_1}\)

or, \(\frac{T_2}{T_1}\) = 1 - \(\frac{30}{100}\) = \(\frac{7}{10}\)

or, T₁ = \(\frac{10T_2}{7}\) = \(\frac{10\times200}{7}\)= 285.71 K

New efficiency is now 50%,

η' = 1 − \(\frac{T_2}{T_1}\)

\(\frac{T_2}{T_1}\) = 1 - η'

\(\frac{T_2}{T_1}\) = 1 - \(\frac{50}{100}\) or \(\frac{1}{2}\)

2T₂ = T₁

or,  T₁ = 2 × 200 K

T₁ = 400 K

Now increase in temperature of source,

= 400 K – 285.71 K

= 114.3 K