To determine the change in internal energy due to the evaporation of one gram of water into steam, we need to consider the latent heat of evaporation and the specific conditions given in the problem. Let's break this down step by step.
Understanding the Concepts
When water evaporates, it undergoes a phase change from liquid to gas. This process requires energy, known as the latent heat of evaporation. In this case, the latent heat is provided as 540 kcal for one gram of water. It's important to note that this energy is absorbed by the water during the phase change, which affects the internal energy of the system.
Key Information
- Mass of water = 1 gram
- Latent heat of evaporation = 540 kcal
- Volume of steam = 1671 cc
- Pressure = 105 Pa
Calculating the Change in Internal Energy
The change in internal energy (ΔU) during the evaporation process can be calculated using the formula:
ΔU = Q - W
Where:
- Q is the heat absorbed (latent heat of evaporation)
- W is the work done by the system during the expansion
Step 1: Heat Absorbed (Q)
In this scenario, the heat absorbed (Q) during the phase change is equal to the latent heat of evaporation:
Q = 540 kcal
Step 2: Work Done (W)
Next, we need to calculate the work done by the steam as it expands against the atmospheric pressure. The work done can be calculated using the formula:
W = PΔV
Where:
- P is the pressure (105 Pa)
- ΔV is the change in volume (final volume of steam - initial volume of water)
Since the initial volume of 1 gram of water is negligible compared to the volume of steam, we can approximate:
ΔV ≈ 1671 cc = 1671 x 10^-6 m³
Now, substituting the values into the work formula:
W = 105 Pa × 1671 x 10^-6 m³
W ≈ 0.175 cal (since 1 J = 0.239 cal, and 1 Pa·m³ = 1 J)
Final Calculation of Change in Internal Energy
Now we can substitute Q and W back into the internal energy equation:
ΔU = Q - W
ΔU = 540 kcal - 0.175 cal
Since 1 kcal = 1000 cal, we convert 540 kcal to calories:
540 kcal = 540,000 cal
Now, substituting this value:
ΔU = 540,000 cal - 0.175 cal ≈ 540,000 cal
Conclusion
Thus, the change in internal energy due to the evaporation of 1 gram of water into steam at 100 °C is approximately 540 cal. Therefore, the correct answer is (B) 540 cal.
