Problem 106 - University Physics with Modern Physics

Question

A tube leads from a \(0.150-\mathrm{kg}\) calorimeter to a flask in which water is boiling under atmospheric pressure. The calorimeter has specific heat capacity 420 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and it originally contains 0.340 \(\mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C}\) . Steam is allowed to condense in the calorimeter at atmospheric pressure until the temperature of the calorimeter and contents reaches \(71.0^{\circ} \mathrm{C}\) , at which point the total mass of the calorimeter and its contents is found to be 0.525 \(\mathrm{kg}\) . Compute the heat of vaporization of water from these data.

Step-by-Step Solution

Answer

The heat of vaporization of water is approximately 2,375,143 J/kg.

Step 1: Identify Initial and Final Mass

The heat problem is based on the change of mass from steam to water. Initially, the mass in the calorimeter is the sum of the mass of the calorimeter and water: \( m_{i} = 0.150 \, \text{kg} + 0.340 \, \text{kg} = 0.490 \, \text{kg} \). After the steam condenses, the final mass is 0.525 kg. Therefore, the mass of the steam that condensed is \( m_{s} = 0.525 \, \text{kg} - 0.490 \, \text{kg} = 0.035 \, \text{kg} \).

Step 2: Determine Heat Absorbed by Water and Calorimeter

Apply the formula for heat absorbed by the calorimeter and the water originally present: \( Q = (m_{water}c_{water} + m_{cal}c_{cal})(T_{f} - T_{i}) \).Substituting the values: \( Q = \left( 0.340 \times 4186 + 0.150 \times 420 \right)(71.0 - 15.0) \).Calculate to find \( Q = (1422 + 63)(56) = 1485 \times 56 = 83160 \, \text{J} \).

Step 3: Relate Heat with Heat of Vaporization

The heat \( Q \) absorbed by the water and calorimeter equals the heat released by the steam as it condenses, given by \( Q = m_{s}L \), where \( L \) is the heat of vaporization of the water.We have \( 83160 = 0.035 \, L \).Thus, \( L = \frac{83160}{0.035} \approx 2,375,143 \, \text{J/kg} \).

Key Concepts

Calorimetry

Calorimetry is a fascinating area of science that deals with measuring the amount of heat exchanged during physical processes. It’s like a detective tool for thermal changes.
When you want to know how much heat is absorbed or released, calorimetry is here to help you. Imagine it's a thermometer that not only records temperature but also calculates heat flow. In our exercise, we're looking at a calorimeter setup.

• The calorimeter is a container – usually insulated – that reduces heat transfer with the environment.
• In the problem, the calorimeter starts with a known mass of water and receives steam from a flask as it reaches equilibrium.

The magic of calorimetry is in its formulas. With the heat equation, we can see how energy moves between steam and water inside the calorimeter. This helps us solve for the heat of vaporization, which is key for converting steam into water through condensation.

Specific Heat Capacity

Specific heat capacity is like a measure of a material's stubbornness in changing temperature! It tells us how much heat is needed to change the temperature of a unit mass by 1 degree Celsius or Kelvin.
In other words, it’s about how easily something can heat up or cool down.
For this exercise:

• The specific heat capacity of water is high (4186 J/kg·K). That means water retains and transfers heat effectively, a vital trait in controlling temperature shifts.
• The calorimeter itself also possesses a specific heat capacity of 420 J/kg·K, calculated separately to understand the heat exchange in myriad parts.

In the context of the calorimetry exercise, knowing these values allows us to find how much the system absorbs until the final temperatures meet. Without these, solving energy problems in calorimetry would feel like chasing wild geese!

Condensation of Steam

Condensation is a process where steam transforms into liquid water, releasing latent heat of vaporization during the change of state. Imagine steam as energetic party-goers with lots of movement, while liquid water is calmer, with less movement.
When steam condenses, it releases a significant amount of this stored energy and slows down.
In our problem:

• When the steam enters the cooler water in the calorimeter, it condenses, warming the water until equilibrium.
• At condensation, heat is released, warming the calorimeter and stored water.

The heat exchanged during condensation helps us determine the heat of vaporization of water. This process is vital in calorimetry, as it’s the bridge that links the steam’s energetic phase to the liquid, heat-storing phase, offering a neat and practical lesson in energy transformation.