Problem 14 - University Physics with Modern Physics

Question

Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is \(2.20 \times 10^{6} \mathrm{J} / \mathrm{kg}\) and the boiling point is \(120^{\circ} \mathrm{C}\) . At this pressure, 1.00 \(\mathrm{kg}\) of water has a volume of \(1.00 \times 10^{-3} \mathrm{m}^{3},\) and 1.00 \(\mathrm{kg}\) of steam has a volume of 0.824 \(\mathrm{m}^{3} .\) (a) Compute the work done when 1.00 \(\mathrm{kg}\) of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

Step-by-Step Solution

Answer

(a) 1.66877 x 10^5 J, (b) 2.033123 x 10^6 J.

Step 1: Understand the Problem

We need to find the work done when 1.00 kg of steam is formed and the increase in internal energy of water at 120°C and 2 atm pressure. We'll need to use the formulae for work done during volume change and the First Law of Thermodynamics.

Step 2: Calculate the Work Done

The formula to calculate work done by a gas during an expansion at constant pressure is \( W = P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the change in volume. Here, \( P = 2.00 \text{ atm} = 2.00 \times 1.013 \times 10^5 \text{ Pa} \), \( V_{f} = 0.824 \text{ m}^3 \), and \( V_i = 1.00 \times 10^{-3} \text{ m}^3 \).Convert atm to Pa and find \( \Delta V = V_{f} - V_i = 0.824 \text{ m}^3 - 1.00 \times 10^{-3} \text{ m}^3 = 0.823 \text{ m}^3 \).Thus, the work done is:\[W = 2.00 \times 1.013 \times 10^5 \text{ Pa} \times 0.823 \text{ m}^3 = 1.66877 \times 10^5 \text{ J}\]

Step 3: Use First Law of Thermodynamics

The First Law of Thermodynamics states \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added, and \( W \) is the work done. Here, \( Q = 2.20 \times 10^6 \text{ J} \) (heat of vaporization for 1 kg).Given the work done \( W = 1.66877 \times 10^5 \text{ J} \), we find the change in internal energy:\[\Delta U = 2.20 \times 10^6 \text{ J} - 1.66877 \times 10^5 \text{ J} = 2.033123 \times 10^6 \text{ J}\]

Step 4: Conclusion

To summarize, the work done when steam is formed is \( 1.66877 \times 10^5 \text{ J} \), and the increase in internal energy of the water is \( 2.033123 \times 10^6 \text{ J} \).

Key Concepts

Boiling Point

The boiling point is the temperature at which a liquid turns into vapor. When we talk about boiling, it's essential to remember that this process occurs when the pressure of the liquid equals the surrounding pressure.
The exercise deals with water boiling at 2.00 atm pressure, leading to a higher boiling point of 120°C compared to the standard 100°C at 1 atm.
Understanding boiling points is crucial because they change with pressure. If you increase the pressure (like in a pressure cooker), the boiling point rises, allowing water to remain a liquid at higher temperatures, which helps cook food faster.
Conversely, in low-pressure environments (like high altitudes), boiling occurs at lower temperatures.

Heat of Vaporization

The heat of vaporization is the energy required to turn a liquid into a gas without changing its temperature. It's like the energy ticket needed for every molecule to break free from the liquid's surface.
In our exercise, the heat of vaporization is given as \(2.20 \times 10^6 \text{ J/kg}\). This means to convert 1 kg of water to steam, you'd need 2.20 million joules of energy.
This is a significant amount because you're overcoming the forces holding the liquid molecules together. That's why evaporating sweat feels cooling – it draws a lot of heat from your skin, using the environment's energy to complete the phase change.

Internal Energy

Internal energy is the total energy contained within a system, determined by the molecular motion and the forces between molecules.
In thermodynamics, a change in internal energy reflects changes in heat content and work done. According to the First Law of Thermodynamics, \(\Delta U = Q - W\). Here, \(Q\) is the heat added, and \(W\) is the work performed by the system.
For this exercise, the added heat is the heat of vaporization, \( 2.20 \times 10^6 \text{ J}\), and the work done, \(1.66877 \times 10^5 \text{ J}\), is subtracted. The result \(\Delta U = 2.033123 \times 10^6 \text{ J}\) tells us how much the internal energy increased as steam formed from liquid water.

Work Done

When a system, like gas, expands against a constant external pressure, mechanical work is done.
The formula for work done during expansion is \(W = P \Delta V\), where \(P\) is pressure, and \(\Delta V\) is the volume change.
In the exercise, water transforms into steam. Its volume increases from \(1.00 \times 10^{-3} \text{ m}^3\) to \(0.824 \text{ m}^3\), with the pressure maintained at \(2.00 \text{ atm} \) (converted to 101300 Pa). Thus, \(\Delta V\) is \(0.823 \text{ m}^3\).
Calculating the work done, we get \(W = 1.66877 \times 10^5 \text{ J}\) as the energy the steam exerts to push against its surrounding. Understanding work in thermodynamics helps us grasp how energy transfer causes movement and changes within a system.