Problem 77 - Chemistry and Chemical Reactivity

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?

Step-by-Step Solution

Answer

Approximately 75.1 g of ice melts.

Step 1: Understand the Heat Transfer

The problem involves transfer of heat from the hot water to the ice, causing the ice to melt and the water to cool until both reach the same temperature, which is the melting point of ice at \(0^{\circ} \text{C}\). We will use the principle of conservation of energy, which states that the heat lost by the water is equal to the heat gained by the ice.

Step 2: Calculate Heat Lost by Water

The amount of heat lost by the water can be calculated using the formula:\[ Q_\text{water} = m_\text{water} \cdot c_\text{water} \cdot \Delta T \]where - \( m_\text{water} = 100.0 \text{ g} \) is the mass of water,- \( c_\text{water} = 4.18 \text{ J/g}^{\circ}\text{C} \) is the specific heat capacity of water,- \( \Delta T = 60.0^{\circ} \text{C} - 0^{\circ} \text{C} = 60.0^{\circ} \text{C}\) is the change in temperature.Plugging in the values:\[ Q_\text{water} = 100.0 \times 4.18 \times 60.0 = 25080 \, \text{J} \]

Step 3: Calculate Heat Needed to Melt Ice

The amount of heat needed to melt a certain mass of ice is given by:\[ Q_\text{ice} = m_\text{ice} \cdot L_f \]where - \( m_\text{ice} \) is the mass of ice melted,- \( L_f = 334 \text{ J/g} \) is the latent heat of fusion of ice.Setting \( Q_\text{ice} \) equal to \( Q_\text{water} \), we have:\[ m_\text{ice} \cdot 334 = 25080 \]

Step 4: Solve for the Mass of Ice Melted

To find the mass of ice that has melted, divide both sides of the equation from Step 3 by \(334\):\[ m_\text{ice} = \frac{25080}{334} \approx 75.1 \text{ g} \]

Step 5: Conclusion

The problem is solved by using the principle of energy conservation to find that approximately \(75.1 \text{ g}\) of ice melts as the system reaches thermal equilibrium at \(0^{\circ} \text{C}\).

Key Concepts

Specific Heat Capacity

Specific heat capacity is a measure of how much heat energy (usually in Joules) is needed to raise the temperature of a certain mass of a substance by a specific amount, typically 1 degree Celsius. It is represented by the symbol \(c\). In this exercise, water has a specific heat capacity of \(4.18 \text{ J/g}^{\circ}{\text{C}}\).
This means that it takes 4.18 Joules of energy to increase the temperature of 1 gram of water by 1 degree Celsius. Understanding this concept is essential as it explains why water, which has a high specific heat capacity, can absorb a lot of heat with only a small change in temperature.

• High specific heat capacity: Can absorb and release large amounts of energy.
• Essential in calculating heat transfer: Used in the formula \(Q = mc\Delta T\), where \(Q\) is the heat exchanged, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

Latent Heat of Fusion

The latent heat of fusion is the amount of heat required to change 1 gram of a solid into a liquid at its melting point without changing temperature. In this exercise, ice has a latent heat of fusion of \(334 \text{ J/g}\).
This concept is crucial when dealing with phase changes, such as when ice melts into water. Even though the temperature stays constant, energy is still needed to break the bonds between ice molecules. This energy does not increase temperature but rather changes the state of the substance.

• Vital for state change: Energy absorbed during melting or released during freezing without temperature change.
• Equation: Used in \(Q = mL_f\), where \(Q\) is the heat involved, \(m\) is the mass, and \(L_f\) is the latent heat of fusion.

Thermal Equilibrium

Thermal equilibrium occurs when two substances, in this case, the water and ice, reach the same temperature and no more heat is transferred between them. This means the system's energy is balanced, resulting in a stable, unchanged state.
In this exercise, once equilibrium is achieved at \(0^{\circ} \text{C}\), no more ice melts, and the water does not cool further. This equilibrium is crucial in calculations because it helps us utilize equations such as the conservation of energy. It applies when the heat lost by the warmer substance is equal to the heat gained by the cooler substance.

• Balance point: Heat from hot object is completely absorbed by cold object, no more net heat flow.
• Foundation for conservation equations: Essential for solving heat transfer problems, assuming total energy exchange equals zero net change.