Question

In a Carnot engine efficiency is 40% at hot reservoir temperature T. For efficiency 50%, what will be the temperature of hot reservoir?
(a) T
(b) 2/3 T
(c) 4/5 T
(d) 6/5 T

Answer 1

percentage efficiency =T(higher temp.)-t(lower)/T*100

(T-t)/T×100=40......(1)

Let K be the new temp.

(K-t)/k×100=50....(2)

from eq 1

T-t=2/5T

t=3/5T put in eq 2

K-3/5T=K/2

K=6/5T