Question

A carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is:

• A. 200°C
• B. 27°C
• C. 15°C
• D. 100°C

Answer 1

The Correct Option is B

To find the temperature of the sink in a Carnot engine, we can use the formula for the efficiency of a Carnot engine, which is given by:

\(\eta = 1 - \frac{T_2}{T_1}\)

where:

• \(\eta\) is the efficiency of the Carnot engine.
• \(T_1\) is the temperature of the source in Kelvin.
• \(T_2\) is the temperature of the sink in Kelvin.

In the given problem, the efficiency \(\eta = 50\% = 0.5\), and the temperature of the source \(T_1 = 327^\circ C = 600 \, \text{K}\).

Substituting the given values into the efficiency equation, we have:

\(0.5 = 1 - \frac{T_2}{600}\)

Rearranging the equation to solve for \(T_2\), we get:

\(\frac{T_2}{600} = 1 - 0.5\)

\(\frac{T_2}{600} = 0.5\)

\(T_2 = 600 \times 0.5\)

\(T_2 = 300 \, \text{K}\)

Now, converting the sink temperature from Kelvin back to Celsius:

\(T_2 = 300 \, \text{K} - 273 = 27^\circ C\)

Therefore, the temperature of the sink is 27°C.

The correct answer is: 27°C

This matches with the provided correct answer. Let's briefly check why the other options are incorrect:

• 200°C: This would imply an exceptionally high sink temperature, leading to lower efficiency than stated.
• 15°C and 100°C: Neither aligns with the calculated result of 27°C when correctly using the Carnot efficiency formula.