Problem 57 - University Physics with Modern Physics

Question

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 into the water to make the final temperature of the system \(40.0^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

0.039 kg of ice is needed.

Step 1: Understand the Problem

You have water at a hot temperature and ice at a cold temperature. You need to find out how much ice should be added so that the final temperature of the entire system is 40.0°C. The system reaches thermal equilibrium where the heat gained by the ice equals the heat lost by the water.

Step 2: Identify Parameters and Formulas

Find the specific heat capacities: water (\( c_w = 4,186 \, \mathrm{J \, kg^{-1} \, \degree C^{-1}} \)) and ice (\( c_i = 2,090 \, \mathrm{J \, kg^{-1} \, \degree C^{-1}} \)). The latent heat of fusion for ice is \( L_f = 334,000 \, \mathrm{J \, kg^{-1}} \). We need the masses of the ice (\( m_{ice} \)) and the final mass of the water is \( m_w = 0.250 \, \mathrm{kg} \). Also note the initial temperatures: \( T_{w} = 75.0 \, \degree C \), \( T_{ice} = -20.0 \, \degree C \), and final temperature \( T_f = 40.0 \, \degree C \).

Step 3: Calculate Heat Lost by Water

The heat lost by the water when it cools down to the final temperature can be calculated using \( Q_{lost} = m_w \, c_w \, (T_w - T_f) \). Substituting values gives \[ Q_{lost} = 0.250 \, \mathrm{kg} \times 4,186 \, \mathrm{J \, kg^{-1} \, \degree C^{-1}} \times (75.0 - 40.0) \, \degree C \]which simplifies to \( Q_{lost} = 36,627.5 \, \mathrm{J} \).

Step 4: Calculate Heat Gained by Ice

The ice will first warm to \(0^{\circ}\mathrm{C}\), then melt, and finally warm to \( 40.0^{\circ}\mathrm{C} \). Calculate it in parts: Part 1: Heating the ice to \(0^{\circ}\mathrm{C} \): \( Q_{1} = m_{ice} \, c_i \, (0 - (-20)) \). Part 2: Melting the ice: \( Q_{2} = m_{ice} \, L_f \). Part 3: Heating the melted water to \( 40.0^{\circ} \mathrm{C} \): \( Q_{3} = m_{ice} \, c_w \, (40 - 0) \).

Step 5: Set Heat Lost Equal to Heat Gained

The total heat the ice gains \( Q_{gain} \) is the sum of \( Q_1 \), \( Q_2 \), and \( Q_3 \). Setting this equal to the heat lost: \[ m_{ice} \, c_i \, (20) + m_{ice} \, L_f + m_{ice} \, c_w \, (40) = 36,627.5 \, \mathrm{J} \] Substitute parameter values:\[ m_{ice} \, (2,090 \, (20) + 334,000 + 4,186 \, (40)) = 36,627.5 \]Solving for \( m_{ice} \) gives: \( m_{ice}=0.039 \mathrm{kg} \).

Step 6: Conclusion

The result tells us that 0.039 kg of ice is needed to reach the final equilibrium temperature of 40.0°C. Ensure to check calculations again for accuracy.

Key Concepts

Thermal Equilibrium

Thermal equilibrium is an important concept when understanding how heat is transferred between objects. It occurs when two or more objects reach the same temperature and, as a result, no more heat flows between them. In the context of the exercise, we consider the water and ice as a single system. This system will reach thermal equilibrium when both the water and the remaining ice (after any changes) stabilize at the same temperature of 40.0°C.

When thermal equilibrium is reached, the heat lost by the warmer object (in this case, the water) is equal to the heat gained by the cooler object (the ice). This balance ensures that the energy within the system is conserved. Utilizing this principle allows us to calculate the amount of ice needed to achieve the desired final temperature within the system.

Specific Heat Capacity

Specific heat capacity is a critical property of materials that tells us how much heat is needed to change the temperature of a substance by a certain amount.
For water, the specific heat capacity is 4,186 J/(kg·°C), and for ice, it is 2,090 J/(kg·°C). This means that it takes more energy to change the temperature of a given mass of water compared to the same mass of ice, due to water's higher specific heat capacity.

Understanding specific heat capacity is essential when calculating how much heat is transferred during a temperature change. In the exercise, we used the specific heat capacities of water and ice to compute the amount of heat lost by the water and gained by the ice at different stages — warming up, melting, and warming again as water.

Latent Heat of Fusion

The latent heat of fusion is the amount of energy required to change a substance from solid to liquid at its melting/freezing point without changing its temperature.
For ice, the latent heat of fusion is 334,000 J/kg. This means that for each kilogram of ice, 334,000 joules of energy are needed to turn it into liquid water at 0°C.

In our scenario, once the ice reaches 0°C, it needs to absorb sufficient heat for it to melt. We include this in our calculation of the total heat gained by the ice. The latent heat of fusion is a vital factor because it represents an additional energy requirement that must be met after the ice reaches its melting point but before it can then increase in temperature as water.

System Temperature Change

The system temperature change involves evaluating how the temperature of the whole setup—water and the ice—adjusts as they interact.
For this exercise, the initial temperature for the water is 75.0°C, and for the ice, it is -20.0°C. Our target is a final system temperature of 40.0°C. To achieve this, we analyze the heat exchanges between the components:

• The water cools from 75.0°C to 40.0°C.
• The ice warms from -20.0°C to 0°C, melts, and then the resulting water warms to 40.0°C.

This process illustrates how energy transfer during temperature changes within a system depends on both heat loss and heat gain. By setting these equal to one another, we can determine how much ice must be used to achieve the desired final temperature.