In Carnot engine, efficiency is 40% at hot reservoir temperature T. For efficiency 50% , what will be the temperature of hot reservoir?

Options

A. `T/5`
B. `(2T)/5`
C. 6T
D. `(6T)/5`

Answer

D

Explanation

`eta =1 -T_(2)/T_(1), T_(1)=T` (Temperature of hot reserviour)
For `eta = 40%`
`= 40/100 =1-T_(2)/T_(1) rArr T_(2)/T = 3/5 rArr T_(2) = 3/5 T`
For `eta = 50%`,
`50/100 = 1 -5/T_(1) rArr T_(1)=(6)/5`