Physics Q15 – NEET-UG 2023

Home Physics ThermodynamicsQuestion #520

A Carnot engine has an efficiency of

50 %

when its source is at a temperature

327^{\circ} C

. The temperature of the sink is :

Clue 1 of 3: Carnot Engine Efficiency

The efficiency of a Carnot engine depends only on the temperatures of the hot source and cold sink.

🥷

Ninja Strategy•Efficiency Estimation

Quickly estimate the efficiency for the higher sink temperatures. Options (3) and (4) would result in efficiencies far below $50 %$ , allowing for their elimination.

🧠 Full Solution Path

Step 1: Convert Source Temperature to Kelvin✦ Active

The given source temperature is $T_{H} = 327^{\circ} C$ . To use it in the Carnot efficiency formula, convert it to Kelvin:

T_{H} = 327 + 273 = 600 K

💡 Teacher's Secret Hint

Always use absolute temperatures (Kelvin) for thermodynamic efficiency calculations.

Step 2: Calculate Sink Temperature in Kelvin○ Expand

The efficiency of a Carnot engine is given by $η = 1 - \frac{T_{C}}{T_{H}}$ . We are given $η = 50 % = 0.5$ and $T_{H} = 600 K$ . Substitute these values to find $T_{C}$ :

0.5 = 1 - \frac{T_{C}}{600}

\frac{T_{C}}{600} = 1 - 0.5

\frac{T_{C}}{600} = 0.5

T_{C} = 0.5 \times 600 = 300 K

Step 3: Convert Sink Temperature to Celsius○ Expand

Convert the sink temperature from Kelvin back to Celsius:

T_{C} (^{\circ} C) = T_{C} (K) - 273

T_{C} (^{\circ} C) = 300 - 273 = 27^{\circ} C

Thus, the temperature of the sink is $27^{\circ} C$ .

💡 Teacher's Secret Hint

Ensure the final answer is in the units requested by the question, which is Celsius in this case.

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Physics Question 15 – NEET-UG 2023

Clue 1 of 3: Carnot Engine Efficiency

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