Question

You add \(100.0 \mathrm{g}\) of water at \(60.0^{\circ} \mathrm{C}\) to \(100.0 \mathrm{g}\) of ice at \(0.00^{\circ} \mathrm{C} .\) Some of the ice melts and cools the water to \(0.00^{\circ} \mathrm{C} .\) When the ice and water mixture reaches thermal equilibrium at \(0^{\circ} \mathrm{C},\) how much ice has melted?

Short answer

75.1 g of ice melts.

Step-by-step solution

Determine the Energy Released by Water

First, calculate the energy released by the water as it cools from \(60.0^{\circ} \mathrm{C} \) to \(0.00^{\circ} \mathrm{C} \). Use the formula:\[Q = mc\Delta T\]where\(m = 100.0 \text{ g} = 0.100 \text{ kg} \), \(c = 4.18 \text{ J/g}^{\circ}\text{C} \), and \( \Delta T = 60.0^{\circ} \mathrm{C} \) (since the final temperature is \(0.00^{\circ} \mathrm{C}\)).Plugging in these values:\[Q = 0.100 \times 4.18 \times 60.0 = 25.08 \text{ kJ}\]So the water releases 25.08 kJ of energy.

Calculate the Energy Required to Melt Ice

Now, calculate how much ice can melt with 25.08 kJ of energy. The energy required to melt ice is calculated using:\[Q_{melting} = m_{ice}\times L_f\]where\(L_f = 334 \text{ J/g} \) (latent heat of fusion for ice).Rearrange the formula to find the mass of ice that melts:\[m_{ice} = \frac{Q_{melting}}{L_f}\]

Find the Mass of Ice Melted

Using the energy \(Q = 25.08\ \text{kJ} = 25080\ \text{J}\):\[m_{ice} = \frac{25080}{334} \approx 75.1\ \text{g}\]So approximately 75.1 g of ice melts.

Key concepts

Heat Transfer

Heat transfer is the process by which thermal energy moves from a hotter object to a colder one. In our example, the 60°C water transfers heat to the 0°C ice. This shift continues until both reach the same temperature, achieving thermal equilibrium.

In this context, we use the formula:

• \( Q = mc\Delta T \)

Here, \( Q \) represents the heat energy transferred, \( m \) is the mass of the water, \( c \) is the specific heat capacity of water, and \( \Delta T \) is the temperature change. The water's heat capacity is 4.18 J/g°C, and it cools from 60°C to 0°C, releasing 25.08 kJ of energy.

This energy then becomes available to potentially melt the ice, further driving the process of heat transfer.

Latent Heat of Fusion

The latent heat of fusion is the amount of energy needed to change a substance from solid to liquid without changing its temperature. For ice, this is \( 334 \text{ J/g} \).

In the exercise, ice absorbs energy from the cooling water. This causes some of the ice to undergo a phase change—transforming from solid ice to liquid water. The temperature of the mixture remains constant at 0°C while this melting occurs.

To determine how much ice melts, use the formula:

• \( Q_{melting} = m_{ice} \times L_f \)

By rearranging for \( m_{ice} \), where \( Q_{melting} \) is the energy available (25.08 kJ) and \( L_f \) is the latent heat of fusion (334 J/g), you calculate the amount of ice melted. In this case, about 75.1 g of ice melts.

Energy Conservation

Energy conservation is a fundamental principle of physics stating that energy cannot be created or destroyed, only transformed. In our problem, the energy the water loses is precisely the energy the ice needs to melt.

When the 100 g of water at 60°C cools to 0°C, it releases 25.08 kJ of energy. This energy isn't lost; instead, it serves to melt part of the ice. Using energy conservation principles, we set:

• Energy lost by water = Energy gained by ice

Thus, conservation allows us to use these energy values to calculate how much ice can melt. Since all energies involved account for the transition to thermal equilibrium, we can predictably calculate the resulting mass of melted ice through the described process.