Question

Efficiency of a Carnot engine is 40% when it receives energy at 500 K. At what temperature it should receive energy to increase its efficiency by 25% :-
(1)
600 K
(2)
700 K
(3)
800 K
(4)
900 K

Answer 1

hello♥️

answer:- 600k

hope it helps ♥️

Answer 2

Dear Student,

◆ Answer - (1)

Ti' = 600 K

● Explanation -

Efficiency of carnot engine is -

η = 1 - To/Ti

At Ti = 500 K,

40/100 = 1 - To/500

0.4 = 1 - To/500

To/500 = 1 - 0.4

To/500 = 0.6

To = 0.6 × 500

To = 300 K

For efficiency to increase by 25%,

40/100 × (100+25)/100 = 1 - 300/Ti'

0.4 × 1.25 = 1 - 300/Ti'

300/Ti' = 1 - 0.5

300/Ti' = 0.5

Ti' = 300/0.5

Ti' = 600 K

Hence, when carnot engine receives energy at 600 K, its efficiency is increased by 25%.

Thanks dear...