Question

An ideal gas heat engine operates in a Carnot cycle between \(227^\circ\text{C}\) and \(127^\circ\text{C}\). It absorbs \(6 \times 10^4 \text{ cal}\) of heat at the higher temperature. The amount of heat converted into work is:

• A. \(1.2 \times 10^4 \text{ cal}\)
• B. \(4.8 \times 10^4 \text{ cal}\)
• C. \(2.4 \times 10^4 \text{ cal}\)
• D. \(6.0 \times 10^3 \text{ cal}\)

Hint:

Always convert Celsius to Kelvin first! Skipping temperature conversions is the most common pitfall in thermodynamics problems. Notice that if you incorrectly used Celsius parameters directly (\(1 - \frac{127}{227}\)), you would end up with an unresolvable fraction that does not match any clean test answer choice.

Answer 1

The Correct Option is A

Concept: The efficiency (\(\eta\)) of a Carnot engine depends exclusively on the absolute thermodynamic temperatures of the hot source (\(T_1\)) and the cold sink (\(T_2\)). It is expressed mathematically as: \[ \eta = 1 - \frac{T_2}{T_1} \] Efficiency also links the total useful mechanical work output (\(W\)) to the heat energy absorbed from the high-temperature reservoir (\(Q_1\)): \[ \eta = \frac{W}{Q_1} \quad \Rightarrow \quad \frac{W}{Q_1} = 1 - \frac{T_2}{T_1} \] Critical Requirement: All temperature inputs must be converted from Celsius to the absolute Kelvin scale (\(T\text{ (K)} = t^\circ\text{C} + 273\)). Step 1: Convert temperatures to the Kelvin scale.

• Source temperature (\(T_1\)): \[ T_1 = 227 + 273 = 500 \text{ K} \]
• Sink temperature (\(T_2\)): \[ T_2 = 127 + 273 = 400 \text{ K} \]

Step 2: Calculate the efficiency (\(\eta\)) of the Carnot engine.
\[ \eta = 1 - \frac{400}{500} = 1 - \frac{4}{5} = \frac{1}{5} = 0.20 \text{ (or } 20\% \text{)} \]

Step 3: Determine the amount of heat converted into work (\(W\)).
Using the relation \(\frac{W}{Q_1} = \eta\), where \(Q_1 = 6 \times 10^4 \text{ cal}\): \[ W = \eta \times Q_1 = \frac{1}{5} \times \left(6 \times 10^4 \right) \] \[ W = 1.2 \times 10^4 \text{ cal} \]