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MCQEasyJEE 2023Second Law & Heat Engines

JEE Physics 2023 Question with Solution

A Carnot engine with efficiency 50%50\% takes heat from a source at 600 K600 \, \text{K}. In order to increase the efficiency to 70%70\%, keeping the temperature of the sink the same, the new temperature of the source will be:

  • A

    360 K360 \, \text{K}

  • B

    1000 K1000 \, \text{K}

  • C

    900 K900 \, \text{K}

  • D

    300 K300 \, \text{K}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Initial efficiency is 50%50\%, source temperature is 600 K600 \, \text{K}, and the sink temperature remains the same.

Find: The new source temperature for efficiency 70%70\%.

A schematic of a Carnot engine between a source at T1 equals 600 K and a sink at T2, with work output W indicated.

For a Carnot engine,

η=1−TsinkTsource\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}

Using the initial efficiency,

0.5=1−Tsink6000.5 = 1 - \frac{T_{\text{sink}}}{600}

So,

Tsink600=0.5\frac{T_{\text{sink}}}{600} = 0.5 Tsink=300 KT_{\text{sink}} = 300 \, \text{K}

Now use the new efficiency 70%70\%:

0.7=1−300Tsource0.7 = 1 - \frac{300}{T_{\text{source}}} 300Tsource=0.3\frac{300}{T_{\text{source}}} = 0.3 Tsource=3000.3=1000 KT_{\text{source}} = \frac{300}{0.3} = 1000 \, \text{K}

Therefore, the new temperature of the source is 1000 K1000 \, \text{K}. The correct option is B. The solution's marks option C, but the worked solution and the listed options show that 1000 K1000 \, \text{K} corresponds to B.

Use sink temperature first

Given: Efficiency changes from 50%50\% to 70%70\%, with initial source temperature 600 K600 \, \text{K}.

Find: New source temperature.

At 50%50\% efficiency, the sink temperature must be half of the source temperature because

η=1−TsinkTsource\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}

So,

Tsink=300 KT_{\text{sink}} = 300 \, \text{K}

For 70%70\% efficiency, the ratio

TsinkTsource=0.3\frac{T_{\text{sink}}}{T_{\text{source}}} = 0.3

Hence,

Tsource=3000.3=1000 KT_{\text{source}} = \frac{300}{0.3} = 1000 \, \text{K}

Therefore, the correct option is B.

Common mistakes

  • Using the Carnot efficiency formula incorrectly as η=1−TsourceTsink\eta = 1 - \frac{T_{\text{source}}}{T_{\text{sink}}}. This reverses the temperature ratio and gives an unphysical result. Always use η=1−TsinkTsource\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}.

  • Treating 50%50\% and 70%70\% as 5050 and 7070 instead of 0.50.5 and 0.70.7. Efficiency must be substituted in fractional form in the equation.

  • Assuming the sink temperature changes in the second case. The question explicitly says the sink temperature remains the same, so first calculate TsinkT_{\text{sink}} from the initial condition and then reuse it.

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