Question

A carnot heat engine has an efficiency of $50\%$ when its sink temperature is ${17}^{o}C$. What must be the change in its source temperature for making efficiency $50\%$:

• A. $120K$
• B. $145K$
• C. $217.5K$
• D. None

Answer 1

Answer: (C)

$217.5K$

Given : A Carnot heat engine has an efficiency of 50% when its sink temperature is ${17}^{o}C$ . What must be the change in its source temperature for making efficiency 50%

Solution :

The final efficiency is 50% of 50% that is 25%

we have formula for efficiency as $\eta =1-\frac{T_2}{T_1}$

$T_1={17}^{o}C$ = 290K

$\eta =\frac{1}{4}=1-\frac{T_2}{290K}$

On Solving we get $T_2=217.5K$