Problem 19 - University Physics with Modern Physics

Question

Boiling Water at High Pressure. When water is boiled at a pressure of \(2.00 \mathrm{atm},\) the heat of vaporization is \(2.20 \times 10^{5} \mathrm{J} / \mathrm{kg}\) 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) Work done is \(1.66 \times 10^5 \mathrm{J}\). (b) Increase in internal energy is \(5.4 \times 10^4 \mathrm{J}\).

Step 1: Calculate Work Done

In thermodynamics, the work done during expansion is given by the formula: \[ W = P \Delta V \]where \( P \) is the pressure and \( \Delta V \) is the change in volume. Here, \( P = 2.00 \, \mathrm{atm} = 2.00 \, \times 1.013 \times 10^5 \, \mathrm{Pa} \). The change in volume, \( \Delta V \), is the final volume of steam minus the initial volume of water: \( \Delta V = 0.824 \, \mathrm{m}^3 - 1.00 \times 10^{-3} \, \mathrm{m}^3 = 0.823 \, \mathrm{m}^3 \). Substituting these values in, we find:\[ W = (2.00 \, \times 1.013 \times 10^5) \times 0.823 = 1.66 \times 10^5 \, \mathrm{J} \]

Step 2: Calculate Change in Internal Energy

The change in internal energy \( \Delta U \) can be found using the first law of thermodynamics:\[ \Delta U = Q - W \]where \( Q \) is the heat added to the system, which is the heat of vaporization \( 2.20 \times 10^5 \, \mathrm{J/kg} \) for 1 kg of water. We have already computed that \( W = 1.66 \times 10^5 \, \mathrm{J} \). Therefore, the change in internal energy is:\[ \Delta U = 2.20 \times 10^5 - 1.66 \times 10^5 = 0.54 \times 10^5 \, \mathrm{J} = 5.4 \times 10^4 \, \mathrm{J} \]

Key Concepts

Heat of vaporization

When water transitions from a liquid to a gas, it undergoes a process known as vaporization. The heat of vaporization is the amount of thermal energy required to convert a certain amount of liquid into a gas at a constant temperature. In this way, it acts as a crucial factor when considering energy changes in a system.
In the context of boiling water at high pressure, such as 2 atm, the energy required to vaporize 1 kg of water becomes essential. The heat of vaporization in our problem is given as \(2.20 \times 10^{5} \text{ J/kg} \).

Key details:

• The heat of vaporization ensures the maintenance of temperature during the phase change.
• Despite the added energy, the temperature remains constant at the boiling point until the phase change is complete.

The understanding of this concept is vital for calculating energy exchanges during boiling under different conditions, such as high pressure.

First law of thermodynamics

The first law of thermodynamics is a pivotal concept in understanding energy relationships in a system and is often reflected as an energy conservation principle. It states that the energy of an isolated system is constant unless altered by an external force. This guiding principle is represented by the equation \( \, \Delta U = Q - W \, \), where \( \, \Delta U \, \) is the change in internal energy, \( \, Q \, \) is the heat added, and \( \, W \, \) is the work done by the system.

In our exercise, this principle is applied to compute the change in internal energy when water boils at a high pressure. Here, water undergoes a state change from liquid to vapor, requiring energy in the form of heat and performing work through expansion.
The first law, therefore, helps us account for:

• The heat energy put into the system (heat of vaporization).
• The energy used in work done due to change in volume under pressure.

This holistic view of energy ensures that we can accurately predict how much internal energy has increased or decreased, reinforcing energy conservation.

Boiling point at high pressure

The boiling point of a liquid is subject to the surrounding atmospheric pressure. At higher pressures, a liquid's boiling point increases.
This is because more energy is needed for molecules to escape into a gaseous state, overcoming the higher external pressure.

In the given problem, boiling occurs at 120°C under 2 atm, unlike the standard 100°C at 1 atm.
Understanding this change in boiling point:

• Helps manipulate and control boiling processes in industrial applications.
• Contributes to more energy-efficient processes by predicting necessary heat inputs at varying pressures.

A higher pressure implies that boiling water can be hotter, storing more thermal energy, which must be accounted for in energy calculations and steam production efficiency.