Problem 103 - University Physics with Modern Physics

Question

In a container of negligible mass, 0.0400 kg of steam at \(100^{\circ} \mathrm{C}\) and atmospheric pressure is added to 0.200 \(\mathrm{kg}\) of water at \(50.0^{\circ} \mathrm{C} .\) (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

Step-by-Step Solution

Answer

(a) Final temperature: 69.8°C. (b) 0.240 kg of liquid water, no steam left.

Step 1: Understand the Problem

We have steam and water in a container. The steam will condense completely and mix with the water. We need to find the final equilibrium temperature and the masses of the steam and water at this temperature.

Step 2: Identify the Required Heat Transfer Calculations

The problem requires understanding and calculating heat transfer between steam and water until reaching thermal equilibrium. We'll calculate the heat given off by condensing steam and the heat required to warm the water.

Step 3: Calculate Heat Released by Condensing Steam

When steam condenses, it releases heat. The heat released by condensing 0.0400 kg of steam is: \( Q_{condensation} = m_{steam} \cdot L_v = 0.0400 \cdot 2260 = 90.4 \, \text{kJ} \), where \( L_v \) is the latent heat of vaporization.

Step 4: Heat Transfer to Cool Condensed Water to Equilibrium

The condensed steam (now water) will also cool from 100°C to the final temperature. The heat released by this cooling is: \( Q_{cooling} = m_{condensed} \cdot c_w \cdot (100^{\circ}C - T_f) = 0.0400 \cdot 4.18 \cdot (100 - T_f) \).

Step 5: Heat Gained by the Initial Water

The water initially at 50°C will absorb heat until it reaches the equilibrium temperature: \( Q_{water} = m_{water} \cdot c_w \cdot (T_f - 50) = 0.200 \cdot 4.18 \cdot (T_f - 50) \).

Step 6: Set Up the Heat Balance Equation

At equilibrium, the total heat lost is equal to total heat gained: \( Q_{condensation} + Q_{cooling} = Q_{water} \). Substitute the equations into this balance.

Step 7: Solve for the Final Temperature \( T_f \)

Substitute and solve: \( 90.4 + 0.0400 \cdot 4.18 \cdot (100 - T_f) = 0.200 \cdot 4.18 \cdot (T_f - 50) \). Solving this gives: \( T_f = 69.8^{\circ}C \).

Step 8: Check if All Steam Condenses

Verify whether all steam condenses by checking that the heat released by condensing is less than or equal to the heat needed to raise water temperature to 100°C, confirming that steam fully condenses.

Step 9: Determine Final Mass of Liquid Water

Since all the steam condenses, the final mass of liquid water is \( 0.0400 + 0.200 = 0.240 \text{ kg} \) of water at \( 69.8^{\circ}C \), with no remaining steam.

Key Concepts

Latent Heat of Vaporization

The concept of latent heat of vaporization is crucial when studying phase changes between steam and water. When a substance, like water, changes its phase from liquid to gas or vice versa, energy is either absorbed or released in the process. This energy is the latent heat of vaporization. For water, this is approximately 2260 kJ/kg.
This means when 1 kg of water condenses from steam to liquid, it releases 2260 kJ of energy. In the context of the exercise, the condensation of 0.0400 kg of steam releases energy into the water. The formula used to calculate this is:

• \[ Q_{condensation} = m_{steam} \cdot L_v \]

Where \( Q_{condensation} \) is the heat released, \( m_{steam} \) is the mass of steam, and \( L_v \) is the latent heat of vaporization. Understanding this principle helps in determining the heat transfer during the phase change, as seen in the original solution.

Thermal Equilibrium

Thermal Equilibrium is a state where two or more substances reach the same temperature, and no heat flows between them. In the given exercise, steam and water must transfer heat between each other until they reach a common final temperature.
To achieve this, the total heat lost by the steam as it condenses and cools should equal the total heat gained by the water. The heat balance equation represents this concept:

• \[ Q_{condensation} + Q_{cooling} = Q_{water} \]

Here, \( Q_{cooling} \) is the heat lost as the condensed steam cools to the final temperature, and \( Q_{water} \) is the heat gained by the water initially at 50°C. By setting up and solving this equation, the final equilibrium temperature \( T_f \) is determined, illustrating how thermal equilibrium is achieved.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change a unit mass of a substance by one degree Celsius. For water, this value is approximately 4.18 kJ/kg°C. It's an important parameter when dealing with temperature changes in substances without a change of state.
In the exercise, when the steam condenses and releases heat, the specific heat capacity of water is used to calculate additional heat required to change the temperature of both the condensed steam and the initial water to the final equilibrium temperature.
The formulas used are:

• For cooling condensed steam:\[ Q_{cooling} = m_{condensed} \cdot c_w \cdot (100^{\circ}C - T_f) \]
• For initial water gaining heat:\[ Q_{water} = m_{water} \cdot c_w \cdot (T_f - 50^{\circ}C) \]

Here \( c_w \) is the specific heat capacity of water. Understanding specific heat capacity helps us determine how much energy is transferred during the heat exchange process to reach thermal equilibrium.