First slide

Heat Engine; Carnot Engine and refrigerator

Question

The efficiency of a carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be?

Moderate

A

100 K
B

400 K
C

600 K
D

500 K

Solution

$\begin{array}{rcl} efficiency of carnot engine is η=1- \frac{T_{2}}{T_{1}} \\ h e r e T_{1} {=temperature of source, T}_{2} =Temperature of sink=500K (given) \\ g i v e n η & = & 50 % = \frac{50}{100} \\ s u b s t i t u t e g i v e n v a l u e s i n e f f i c i e n c y e q u a t i o n \\ \frac{1}{2} & = & 1 - \frac{500}{T_{1}} \\ T_{1} =1000 K \\ B y t h e q u e s t i o n \\ if η=60%= \frac{60}{100} \\ η =1- \frac{T_{2}}{T_{1}} \\ s u b s t i t u t e t h e g i v e n v a l u e s i n t h e s e c o n d p a r t o f q u e s t i o n \\ \frac{60}{100} =1- \frac{T_{2}}{1000} \\ \frac{T_{2}}{1000} & = & 1 - \frac{6}{10} \\ T_{2} =400 K \\ t e m p e r a t u r e o f \sin k =400 K \end{array}$

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