a carnot engine whose sink is at 90 k has an efficiency of 30%. by how much the teperature of the source be increased to have its efficiency equal to 50%, keeping sink teperature constant

Question

Khimraj · Answer

efficiency = 1 – T sink /T source +1 = T sink /T source As T source is constant ( 2 +1)/( 1 ​ +1) = T sink2 /T sink1 1.6/1.3 = x/90 x = 110.76 %increase in teperature of the source = 20.76*100/90 = 23% Hope it clears.....................................

Aastha Singh · Answer

DEAR STUDENT , KINDLY REFER TO THE IMAGE BELOW

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

a carnot engine whose sink is at 90 k has an efficiency of 30%. by how much the teperature of the source be increased to have its efficiency equal to 50%, keeping sink teperature constant

a carnot engine whose sink is at 90 k has an efficiency of 30%. by how much the teperature of the source be increased to have its efficiency equal to 50%, keeping sink teperature constant

2 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Thermal Physics

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship