Question

A ideal carnot engine whose efficiency is 40% receives heat at 500k.if it's efficiency is 50% then the intake temperature for the same exaust temperature

Answer 1

Answer:

600k

Explanation:

In this,efficiency formula is used.

In 1st T1 is taken as 500K.

nd in 2nd T2 is taken as 300K.

Answer 2

Dear Student,

◆ Answer -

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 be 50%,

50/100 = 1 - 300/Ti'

0.5 = 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 50%.

Thanks dear...