Question

A Carnot engine takes 5000 kcal of heat from a reservoir at 727°C and gives heat to a sink at 127°C. The work done by the engine is

• A. \(3 × 10^6\) J
• B. Zero
• C. \(12.6 × 10^6\) J
• D. \(8.4 × 10^6\) J

Answer 1

The Correct Option is C

 To determine the work done by the Carnot engine, we need to follow these steps:

1. Convert the given temperatures from Celsius to Kelvin, as thermodynamic calculations require absolute temperatures.
   • Temperature of the hot reservoir: \(T_1 = 727^{\circ} \text{C} = 727 + 273 = 1000 \, \text{K}\)
   • Temperature of the cold reservoir: \(T_2 = 127^{\circ} \text{C} = 127 + 273 = 400 \, \text{K}\)
2. Recall the Carnot engine efficiency formula: \(\eta = 1 - \frac{T_2}{T_1}\)
   • Substitute the values: \(\eta = 1 - \frac{400}{1000} = 1 - 0.4 = 0.6\)
3. Calculate the work done by using the relation: \(\text{Work Done (W)} = \eta \times Q_1\)
   • Given heat taken from the reservoir, \(Q_1 = 5000 \, \text{kcal}\), convert it to Joules:
   • \(Q_1 = 5000 \times 4184 \, \text{J} = 20920000 \, \text{J}\)
   • Calculate the work done:
   • \(W = 0.6 \times 20920000 \, \text{J} = 12552000 \, \text{J}\)
   • The closest option to this value is given as \(12.6 \times 10^6\) J because the exact multiplication of \(0.6 \times 20920000\) is \(12552000\) which can be rounded and represented as \(12.6 \times 10^6\) J.

Therefore, the work done by the engine is \(12.6 \times 10^6\) J.