Question

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

Answer 1

Efficiency n = (t2-t1)/ t2 

Where t2 is the source temperature and t1 is the sink temperature  

Now when the efficiency is 50% 

0.5 = (t2-500)/T2 

T2 = 1000K 

When the efficiency is  

0.6 = (t2-t1)/T2 

The required temperature of the sink will be  

T1 = 400k

Answer 2

Explanation:

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