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JEE · NEET Previous Year Question

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:

Topic: NEET PYQ

Answer: C. Full step-by-step solution below.

  1. A600 K
  2. B500 K
  3. C400 K
  4. D100 K

Correct answer: C

Solution

(c) Efficiency of the Carnot engine is given by
η = 1 –
where T1 = temperature of source
T2 = temperature of sink
Given η = 50% = 0.5, T2 = 500 K
Substituting in relation (i), we have
0.5 = 1 –
or
∴ T1 = = 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 T2 = 0.4 × 1000 = 400 K

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