The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If the temperature of source is kept constant and its efficiency is to be raised to 60%; then the required temperature of the sink will be:
Text Solution
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Efficiency of the Carnot engine is given by
η = 1 – 
where T 1 = temperature of source
T 2 = temperature of sink
Given η = 50% = 0.5, T2 = 500 K
Substituting in relation (i), we have
0.5 = 1 – 
or 
∴ T 1 = 
= 1000 K
Now, the temperature of sink is changed to T2 and the efficiency becomes 60% i.e., 0.6.
Using relation (i), we get
0.6 = 1 – 
or 
= 1 – 0.6 = 0.4
or T 2 = 0.4 × 1000 = 400 K
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