Problem 48 - College Physics

Question

\(\bullet\) An insulated beaker with negligible mass contains 0.250 \(\mathrm{kg}\) of water at a temperature of \(75.0^{\circ} \mathrm{C}\) . How many kilograms of ice at a temperature of \(-20.0^{\circ} \mathrm{C}\) must be dropped in the water so that the final temperature of the system will be \(30.0^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

Approximately 0.128 kg of ice is required.

Step 1: Understand the problem

We need to determine the mass of ice required to cool down water from \(75^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\), while the ice warms up from \(-20^{\circ} \mathrm{C}\) and eventually melts.

Step 2: Define the energy changes

The problem involves three energy changes: 1. Heating the ice from \(-20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\). 2. Melting the ice at \(0^{\circ} \mathrm{C}\). 3. Mixing the resulting water from melted ice with the initial water, to reach the final temperature of \(30^{\circ} \mathrm{C}\).

Step 3: Calculate energy to heat ice

The specific heat capacity of ice is \(c_{\text{ice}} = 2.09 \, \mathrm{J/g^{\circ}C}\). The heat required to raise the temperature of the ice from \(-20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) is calculated as: \[ Q_1 = m_{\text{ice}} \times c_{\text{ice}} \times (0 - (-20)) = m_{\text{ice}} \times 2.09 \times 20 \]

Step 4: Calculate energy to melt ice

The latent heat of fusion of ice is \(L_f = 334 \, \mathrm{J/g}\). Thus, the energy required to turn the ice at \(0^{\circ} \mathrm{C}\) into water at \(0^{\circ} \mathrm{C}\) is: \[ Q_2 = m_{\text{ice}} \times L_f = m_{\text{ice}} \times 334 \]

Step 5: Calculate energy to mix water to final temperature

After ice becomes water at \(0^{\circ} \mathrm{C}\), it is heated to \(30^{\circ} \mathrm{C}\). The specific heat capacity of water is \(c_{\text{water}} = 4.18 \, \mathrm{J/g^{\circ}C}\). So the heat needed is: \[ Q_3 = m_{\text{ice}} \times c_{\text{water}} \times (30 - 0) = m_{\text{ice}} \times 4.18 \times 30 \]

Step 6: Calculate energy released by water cooling

The water cools from \(75^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\). The heat released is: \[ Q_4 = m_{\text{water}} \times c_{\text{water}} \times (75 - 30) = 250 \times 4.18 \times 45 \] Convert mass to grams as needed.

Step 7: Set up the energy balance equation

According to the conservation of energy, the heat lost by water equals the heat gained by ice: \[ Q_4 = Q_1 + Q_2 + Q_3 \] Substitute the expressions from the previous steps and solve for \(m_{\text{ice}}\).

Step 8: Solve for ice mass

Insert the values from the equations: \[ 250 \times 4.18 \times 45 = m_{\text{ice}} \times 2.09 \times 20 + m_{\text{ice}} \times 334 + m_{\text{ice}} \times 4.18 \times 30 \] Simplify: \[ m_{\text{ice}} (2.09 \times 20 + 334 + 4.18 \times 30) \approx 47025 \text{ J} \] Calculate \(m_{\text{ice}}\).

Step 9: Conclusion

Solving the equation gives \(m_{\text{ice}} \approx 0.128 \, \mathrm{kg}\). This is the mass of ice required to cool the system to \(30^{\circ} \mathrm{C}\).

Key Concepts

Heat Transfer

Heat transfer is the process of energy moving from one system to another due to a temperature difference. In our scenario, heat transfer occurs between the hot water and the ice. The heat moves from the warmer water, which is initially at 75°C, to the colder ice, which starts at -20°C.

There are a few key mechanisms through which heat transfer can occur:

• Conduction: Energy transfer through direct contact. Here, heat transfers when ice and water come into contact.
• Convection: Energy transfer through fluid motion. In liquids and gases, the warmer segments tend to rise while cooler ones sink, aiding in heat exchange.
• Radiation: Energy transfer through electromagnetic waves. However, this is not significant in this specific scenario.

Understanding heat transfer is crucial in predicting temperature changes in thermodynamics.

Specific Heat Capacity

Specific heat capacity is a property that describes how much heat energy a substance can absorb per unit mass per degree change in temperature. It tells us how substances react to heating or cooling.

In our case,

• For ice, the specific heat capacity is 2.09 J/g°C.
• For water, the specific heat capacity is 4.18 J/g°C.

When applying this concept to our exercise, specific heat capacity helps in calculating the energy change when adjusting the temperature of water and ice. For example, to heat 1 kilogram of ice by 1°C, you'd need 2.09 kJ of energy. This number increases to 4.18 kJ per kilogram per degree for water due to its higher specific heat capacity. Its importance is fundamental to understanding how substances absorb and conserve energy differently.

Latent Heat

Latent heat is the amount of heat absorbed or released by a substance during a phase change without changing temperature. During a phase transition, such as melting or freezing, the substance absorbs or releases a significant amount of heat, known as latent heat.

In this exercise, we consider the latent heat of fusion for ice, which is the energy needed to change ice from solid to liquid without changing its temperature. For ice, this value is 334 J/g.

• This means turning 1 gram of ice to water at 0°C requires 334 joules.
• It's a critical energy component in solving how much ice is needed to achieve a final water temperature of 30°C in the system.

By including latent heat in our calculations, we ensure the change in states doesn't skew our temperature balance predictions, conserving overall energy in the process.

Energy Conservation

Energy conservation is a fundamental principle of physics stating that energy cannot be created or destroyed, only transferred or converted from one form to another. In thermodynamic systems like our exercise, energy transfer happens between the ice and water, maintaining this principle.

The exercise draws a direct application of energy conservation:

• The heat lost by the water as it cools down from 75°C to 30°C is precisely the heat gained by the ice as it warms, melts, and then increases in temperature to the final equilibrium state.
• This energy balance is described mathematically in the equation: \(Q_4 = Q_1 + Q_2 + Q_3\), where each \(Q\) represents different stages of heat absorption or release.

Understanding energy conservation helps solve problems involving heat, ensuring that the calculated change in energy matches the gained and lost energy within the system.