Question

An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 x 10⁴ cal of heat at higher temperature. Amount of heat converted to work is:

• A. 2.4 x 10⁴ cal
• B. 3.6 x 10⁴ ca
• C. 1.2 x 10⁴ cal
• D. 6.4 x 10⁴ cal

Answer 1

The Correct Option is C

To solve this problem, we must understand the principles of a Carnot cycle and how it operates as a heat engine.

The Carnot cycle is an idealized thermodynamic cycle that provides the maximum possible efficiency for a heat engine operating between two temperatures.

1. First, let's convert the given temperatures from Celsius to Kelvin, as thermodynamic calculations require absolute temperatures. T_1 = 227^\circ C + 273 = 500 \, \text{K}, and T_2 = 127^\circ C + 273 = 400 \, \text{K}.
2. The efficiency (\eta) of a Carnot engine is given by the formula: \eta = 1 - \frac{T_2}{T_1}.
3. Substituting the values: \eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2 or 20%.
4. The work done by the engine is the fraction of the absorbed heat that is converted into work, given by W = \eta Q_1, where Q_1 is the heat absorbed.
5. Given Q_1 = 6 \times 10^4 \, \text{cal}: W = 0.2 \times 6 \times 10^4 = 1.2 \times 10^4 \, \text{cal}.

This calculation confirms that the amount of heat converted to work by the Carnot engine is 1.2 \times 10^4 \, \text{cal}.

Therefore, the correct answer is 1.2 x 10⁴ cal.