To solve this problem, let's go step by step. We need to find:

1. The work done by the system ($W$)
2. The heat transferred to the system ($Q$)
3. The change in internal energy ($\Delta U$)

We'll be using the following thermodynamic relations:

• First Law of Thermodynamics: $\Delta U = Q - W$
• Work Done at Constant Pressure: $W = P \Delta V$ where:
• $\Delta U$ is the change in internal energy,
• $Q$ is the heat transferred,
• $W$ is the work done,
• $P$ is the pressure, and
• $\Delta V$ is the change in volume.

Step 1: Work Done ($W$)

We know:

• Pressure $P = 1$ atm = $1.013 \times 10^5$ Pa (since $1 \text{ atm} = 101325 \text{ Pa}$)
• Mass of water, $m = 1.0$ kg
• Molar mass of water, $M = 18.015$ g/mol

The specific volume change ($\Delta V$) can be calculated using the molar volume change from water to steam:

• Initial state: Water (liquid) at 100°C
• Final state: Steam (vapor) at 100°C

The molar volume of liquid water at 100°C is approximately $1 \times 10^{-3} \text{ m}^3/\text{kg}$. The molar volume of water vapor at 100°C and 1 atm can be calculated using the ideal gas law: $V_m = \frac{RT}{P}$ where:

• $R = 8.314$ J/(mol·K) is the universal gas constant,
• $T = 373.15$ K (temperature in Kelvin),
• $P = 1.013 \times 10^5$ Pa.

$V_m = \frac{8.314 \times 373.15}{1.013 \times 10^5} = 0.0306 \text{ m}^3/\text{mol}$

Now convert this to volume per kg: $V_{\text{steam}} = \frac{V_m}{M} = \frac{0.0306}{0.018015} = 1.7 \text{ m}^3/\text{kg}$

Change in Volume

$\Delta V = V_{\text{steam}} - V_{\text{water}} = 1.7 \text{ m}^3 - 0.001 \text{ m}^3 = 1.699 \text{ m}^3$

Now we can calculate the work done: $W = P \Delta V = (1.013 \times 10^5 \text{ Pa}) \times 1.699 \text{ m}^3 = 1.72 \times 10^5 \text{ J}$

Step 2: Heat Transferred ($Q$)

The heat required to convert water to steam at 100°C is given by the latent heat of vaporization ($L$): $Q = mL$ where:

• $L = 2260 \text{ kJ/kg} = 2.26 \times 10^6 \text{ J/kg}$ (latent heat of vaporization at 100°C) $Q = 1 \text{ kg} \times 2.26 \times 10^6 \text{ J/kg} = 2.26 \times 10^6 \text{ J}$

Step 3: Change in Internal Energy ($\Delta U$)

Using the first law of thermodynamics: $\Delta U = Q - W = 2.26 \times 10^6 \text{ J} - 1.72 \times 10^5 \text{ J}$ $\Delta U = 2.088 \times 10^6 \text{ J}$

Summary

1. Work Done by the System ($W$): $1.72 \times 10^5 \text{ J}$
2. Energy Transferred as Heat ($Q$): $2.26 \times 10^6 \text{ J}$
3. Change in Internal Energy ($\Delta U$): $2.088 \times 10^6 \text{ J}$

Would you like more details or have any questions?

Further Questions

1. What is the relationship between internal energy change and heat in an isochoric process?
2. How can the specific heat capacity affect the amount of heat required for phase change?
3. How does the ideal gas law help in calculating the volume of water vapor?
4. Can you explain the importance of latent heat in phase transitions?
5. What happens to the work done if the pressure is not constant?

Tip:

Always ensure that units are consistent throughout thermodynamic calculations to avoid errors.