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

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

Regards

Arun (askIITians forum expert)