Problem 76 - Chemistry and Chemical Reactivity

Question

Three 45 -g ice cubes at \(0^{\circ} \mathrm{C}\) are dropped into \(5.00 \times\) \(10^{2} \mathrm{mL}\) of tea to make ice tea. The tea was initially at \(20.0^{\circ} \mathrm{C} ;\) when thermal equilibrium was reached, the final temperature was \(0^{\circ} \mathrm{C}\). How much of the ice melted and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water.

Step-by-Step Solution

Answer

125.15 g of ice melted, 9.85 g remained.

Step 1: Calculate Heat Lost by Tea

Start by determining the heat lost by the tea to reach the equilibrium temperature of \(0^{\circ} \mathrm{C}\). Use the formula for heat transfer: \[ q = mc\Delta T \]where:- \( m = 500 \text{ g} \) (equivalent to \( 500 \text{ mL} \) since the density of water/tea is approximately \( 1 \text{ g/mL} \)), - \( c = 4.18 \text{ J/g}^\circ \text{C} \) (specific heat capacity of water),- \( \Delta T = 20.0^{\circ} \mathrm{C} \) (change in temperature).Thus, \[ q = 500 \text{ g} \times 4.18 \text{ J/g}^\circ \text{C} \times 20.0^{\circ} \mathrm{C} = 41,800 \text{ J} \]

Step 2: Determine Amount of Ice Melted

Use the heat absorbed by the ice to calculate how much ice melted. The heat absorbed is equal to the heat lost by the tea, which is 41,800 J.The formula for heat absorbed by ice to melt is:\[ q = mL_f \]where:- \( m \) is the mass of ice melted,- \( L_f = 334 \text{ J/g} \) (latent heat of fusion of ice).Rearranging gives:\[ m = \frac{q}{L_f} = \frac{41,800 \text{ J}}{334 \text{ J/g}} = 125.15 \text{ g} \]

Step 3: Calculate Remaining Ice

Calculate the total initial mass of ice and compare it with the amount that melted to find the ice that remained.Total initial mass of ice: \[ 3 \times 45 \text{ g} = 135 \text{ g} \]Amount that remained is:\[ 135 \text{ g} - 125.15 \text{ g} = 9.85 \text{ g} \]

Step 4: Conclusion

The ice melted until thermal equilibrium was reached at \(0^{\circ} \mathrm{C}\), with 125.15 g of ice having melted. Therefore, 9.85 g of ice remained floating in the tea.

Key Concepts

Heat Transfer

Heat transfer is a fundamental concept that explains how thermal energy moves from one object to another. This transfer occurs naturally from a warmer object to a cooler one until thermal equilibrium is reached, meaning both objects are at the same temperature. In our exercise, heat transfer happens when the hot tea at 20°C is cooled down by the ice cubes until both reach a final temperature of 0°C.
The heat lost by the tea is calculated using the formula: \[ q = mc\Delta T \]where:

• \( q \) is the heat transfer in joules (J),
• \( m \) is the mass of the tea in grams (g),
• \( c \) is the specific heat capacity,
• \( \Delta T \) is the change in temperature.

Here, the tea loses 41,800 J of thermal energy as it cools from 20°C to 0°C.

Specific Heat Capacity

Specific heat capacity is a property of matter that indicates how much heat energy is needed to raise the temperature of one gram of a substance by one degree Celsius. It varies across different materials and plays a crucial role in understanding heat transfer. In the context of our exercise, we consider the specific heat capacity of tea to be the same as that of pure water, which is 4.18 J/g°C.
This means that to change the temperature of one gram of tea by one degree Celsius, 4.18 joules of energy are required. Knowing this helps us calculate the total amount of heat lost by the tea when cooling down, aiding in determining how much ice melted.

Latent Heat of Fusion

Latent heat of fusion is the amount of heat required to change a unit mass of a solid into a liquid without a change in temperature. For ice, this means turning solid ice into liquid water while remaining at 0°C. For this concept, we use the formula:\[ q = mL_f \]where:

• \( q \) is the heat absorbed or released (in this case absorbed),
• \( m \) is the mass of the ice melted,
• \( L_f \) is the latent heat of fusion, which is 334 J/g for ice.

During the exercise, the heat absorbed by the ice from the tea was used to calculate the amount of ice that melted – which was 125.15 g. Knowing this value helps in understanding how much energy is required to transition the ice into liquid form without altering its temperature.