Problem 80 - Physics

Question

A \(0.200-\mathrm{kg}\) piece of aluminum that has a temperature of \(-155^{\circ} \mathrm{C}\) is added to \(1.5 \mathrm{~kg}\) of water that has a temperature of \(3.0^{\circ} \mathrm{C}\). At equilibrium the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

Step-by-Step Solution

Answer

Approximately 0.027 kg of water freezes into ice.

Step 1: Understand the Problem

We need to calculate how much water at \(3.0^{\circ} \mathrm{C}\) freezes when a very cold piece of aluminum is added to it. The final equilibrium temperature is \(0.0^{\circ} \mathrm{C}\). This means some of the water has turned into ice because it's the only way to absorb more heat than just cooling to \(0.0^{\circ} \mathrm{C}\).

Step 2: List Given Information and Constants

- Mass of aluminum \( (m_{\text{Al}}) = 0.200 \, \text{kg} \)- Mass of water initially \( (m_{\text{water, initial}}) = 1.5 \, \text{kg} \)- Initial temperature of aluminum \( = -155^{\circ} \mathrm{C} \)- Initial temperature of water \( = 3^{\circ} \mathrm{C} \)- Equilibrium temperature \( = 0^{\circ} \mathrm{C} \)- Specific heat of aluminum \( (c_{\text{Al}}) = 900 \, \mathrm{J/kg}^{\circ} \mathrm{C} \)- Specific heat of water \( (c_{\text{water}}) = 4186 \, \mathrm{J/kg}^{\circ} \mathrm{C} \)- Latent heat of fusion for water \( (L_f) = 334,000 \, \mathrm{J/kg} \)

Step 3: Calculate Heat Lost by Water

Calculate the heat lost by the water when cooling from \(3^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\):\[ Q_{\text{water, cooling}} = m_{\text{water, initial}} \cdot c_{\text{water}} \cdot (0^{\circ} \mathrm{C} - 3^{\circ} \mathrm{C}) \]\[ Q_{\text{water, cooling}} = 1.5 \, \text{kg} \times 4186 \, \mathrm{J/kg^{\circ} C} \times (-3^{\circ}) = -18,837 \, \mathrm{J} \]

Step 4: Calculate Heat Gained by Aluminum

Calculate the heat gained by the aluminum as it warms from \(-155^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\):\[ Q_{\text{Al}} = m_{\text{Al}} \cdot c_{\text{Al}} \cdot (0^{\circ} \mathrm{C} - (-155^{\circ} \mathrm{C})) \]\[ Q_{\text{Al}} = 0.200 \, \text{kg} \times 900 \, \mathrm{J/kg^{\circ} C} \times 155^{\circ} = 27,900 \, \mathrm{J} \]

Step 5: Determine Additional Heat Absorbed by Aluminum

To reach equilibrium, more heat must be absorbed by aluminum than just to warm it up due to water freezing. Calculate the heat that goes to freeze the water.The difference in heat between what aluminum gained and what water lost is absorbed through the freezing process:\[ Q_{\text{freeze}} = Q_{\text{Al}} - |Q_{\text{water, cooling}}| \]\[ Q_{\text{freeze}} = 27,900 \, \mathrm{J} - 18,837 \, \mathrm{J} = 9,063 \, \mathrm{J} \]

Step 6: Calculate Mass of Water Frozen

Calculate the mass of water frozen using the latent heat of fusion:\[ m_{\text{frozen}} = \frac{Q_{\text{freeze}}}{L_f} \]\[ m_{\text{frozen}} = \frac{9,063 \, \mathrm{J}}{334,000 \, \mathrm{J/kg}} \approx 0.027 \, \mathrm{kg} \]

Step 7: Solution Response

The mass of water that has been frozen into ice is approximately \(0.027 \, \mathrm{kg}\).

Key Concepts

Heat Transfer

In this exercise, we explore the concept of heat transfer, which is a fundamental principle of thermodynamics, the branch of physics that deals with heat and temperature interactions. Heat transfer is the process by which thermal energy is exchanged between physical systems. Here, the very cold aluminum loses heat to the relatively warmer water. This energy transfer occurs until both materials reach the same temperature — called thermal equilibrium.

Heat always moves from a warmer object to a cooler object. In our example, the water starts at a higher temperature than the aluminum. As a result, heat flows from the water to the aluminum. This causes a temperature change in both substances, and is a classic demonstration of how energy is distributed between different materials, leading to changes in their physical states.

Latent Heat of Fusion

The latent heat of fusion is a key factor when a substance changes its state from solid to liquid or vice versa. In this exercise, part of the water freezes into ice. The latent heat of fusion denotes the amount of heat needed to change a given quantity of substance from liquid to solid without changing its temperature.

For water, this value is substantial because a significant amount of energy is involved in the phase change even though the temperature doesn't change during the process. The latent heat of fusion for water is crucial to calculate the mass of the water that transitions into ice. It helps understand how much energy the aluminum can absorb from the water, thus leading some of it to freeze.

Specific Heat Capacity

Specific heat capacity is another essential concept in this exercise. It refers to the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. Different materials have different specific heat capacities.

Aluminum, with a specific heat capacity of 900 J/kg°C, heats up quickly but also cools quickly. On the other hand, water has a specific heat capacity of 4186 J/kg°C, showing its ability to store more heat compared to aluminum. In the problem, these values allow us to calculate how much energy is transferred between the water and the aluminum. This in turn influences the equilibrium temperature, facilitating the calculations involved when determining the amount of water that turns to ice.

Equilibrium Temperature

The equilibrium temperature in thermodynamics is the point at which the substances in contact have exchanged heat until there is no net heat flow between them. In the exercise, both the aluminum and the water reached an equilibrium temperature of 0°C. This means that the energy lost by the water equaled the energy gained by the aluminum.

This concept is essential because it determines the state of the substances involved (whether some of the water turns into ice, for example). Equilibrium plays a critical role in calculations, as it allows the determination of how much heat was transferred. Thus, it provides insights into how and when phase changes result from thermal interactions.