To solve this problem, we can apply the principles of a Carnot engine, which operates between two heat reservoirs with temperatures $T_{\text{cold}}$ and $T_{\text{hot}}$.

Given Data:

• $Q_{\text{in}} = 5 \, \text{kJ}$ (heat extracted from the cold reservoir)
• $T_{\text{cold}} = -10^{\circ} \text{C} = 263 \, \text{K}$
• $T_{\text{hot}} = 25^{\circ} \text{C} = 298 \, \text{K}$

Carnot Efficiency Formula

For a Carnot engine, the efficiency $\eta$ is given by: $\eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$

Substitute the values: $\eta = 1 - \frac{263}{298}$ $\eta \approx 0.1174 \, \text{or} \, 11.74\%$

Work Done (W)

The efficiency of the Carnot engine is also defined as the ratio of the work output $W$ to the heat input $Q_{\text{in}}$: $\eta = \frac{W}{Q_{\text{in}}}$ So, we can solve for $W$: $W = \eta \cdot Q_{\text{in}}$ $W = 0.1174 \times 5 \, \text{kJ}$ $W \approx 0.587 \, \text{kJ}$

Final Answer

The work done by the Carnot device is approximately 0.587 kJ.

Would you like further details or have any questions?

Here are some additional questions to expand your understanding:

1. What happens to the efficiency if the temperatures of the reservoirs increase?
2. How does a real engine's efficiency compare to the Carnot efficiency?
3. What are other practical examples of devices that use the Carnot cycle?
4. How does the work output change if the temperature difference between reservoirs is larger?
5. Can a Carnot cycle operate if both reservoirs are at the same temperature?

Tip: In thermodynamics, higher efficiency is achieved when the temperature difference between the heat source and sink is larger.