Question

In a Carnot engine, the heat released to the sink is \(22\%\) less than the heat taken from the source. If the temperature of the source is \(127^\circ\text{C}\), then the temperature of the sink is:

• A. \(78^\circ\text{C}\)
• B. \(39^\circ\text{C}\)
• C. \(27^\circ\text{C}\)
• D. \(22^\circ\text{C}\)

Hint:

Always convert Celsius temperatures into Kelvin before applying Carnot engine relations.

Answer 1

The Correct Option is B

Concept: For a Carnot engine, \[ \frac{Q_2}{Q_1}=\frac{T_2}{T_1} \] where \[ Q_1=\text{heat absorbed} \] \[ Q_2=\text{heat rejected} \] \[ T_1=\text{source temperature} \] \[ T_2=\text{sink temperature} \]

Step 1: Express the rejected heat in terms of absorbed heat. Heat rejected is \(22\%\) less than heat absorbed. \[ Q_2=0.78Q_1 \] Thus, \[ \frac{Q_2}{Q_1}=0.78 \]

Step 2: Convert source temperature to Kelvin. \[ T_1=127+273 \] \[ T_1=400K \]

Step 3: Find sink temperature. \[ \frac{T_2}{400}=0.78 \] \[ T_2=312K \] \[ T_2=312-273 \] \[ T_2=39^\circ C \]

Step 4: Final answer. \[ \boxed{39^\circ C} \]