Problem 107 - Chemistry The Central Science In Si Units

Question

An ice cube with a mass of \(25 \mathrm{~g}\) at \(-18^{\circ} \mathrm{C}\) (typical freezer temperature) is dropped into a cup that holds \(250 \mathrm{~mL}\) of hot water, initially at \(85^{\circ} \mathrm{C}\). What is the final temperature in the cup? The density of liquid water is \(1.00 \mathrm{~g} / \mathrm{mL} ;\) the specific heat capacity of ice is \(2.03 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} ;\) the specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}-\mathrm{K} ;\) the enthalpy of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\).

Step-by-Step Solution

Answer

The final temperature in the cup is approximately \(58.6^{\circ} \mathrm{C}\).

Step 1: Calculate the Heat Required to Warm the Ice to 0°C

To warm the ice from \(-18^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\), we use the formula \(q = mc\Delta T\). For the ice, \(m = 25 \mathrm{~g}\), \(c = 2.03 \mathrm{~J/g}^{\circ}\mathrm{C}\), and \(\Delta T = 0 - (-18) = 18^{\circ} \mathrm{C}\). Thus, the heat required is \(q = 25 \times 2.03 \times 18 = 914.5 \mathrm{~J}\).

Step 2: Calculate the Heat Required to Melt the Ice

To melt the ice, we use the enthalpy of fusion. The molar mass of water is \(18 \mathrm{~g/mol}\), so the moles of ice are \(\frac{25}{18}\). The enthalpy of fusion is \(6.01 \mathrm{~kJ/mol}\), which converts to \(6010 \mathrm{~J/mol}\). The heat required is \(\frac{25}{18} \times 6010 = 8347.2 \mathrm{~J}\).

Step 3: Calculate the Heat Loss by the Hot Water

The mass of the hot water is \(250 \mathrm{~g}\) (since \(1 \mathrm{~mL} = 1 \mathrm{~g}\)). The water cools from \(85^{\circ} \mathrm{C}\) to some final temperature \(T_f\), using the formula \(q = mc\Delta T\). Thus, \(q = 250 \times 4.184 \times (85 - T_f)\).

Step 4: Set up and Solve the Equation for Final Temperature

The heat gained by the ice must equal the heat lost by the water. So, \(914.5 + 8347.2 + 25 \times 4.184 \times (T_f - 0) = 250 \times 4.184 \times (85 - T_f)\). Simplifying, this becomes \(5735T_f = 86360 - 9261.2\). Solving for \(T_f\), we find \(T_f \approx 58.6^{\circ} \mathrm{C}\).

Step 5: Verify the Physical Feasibility

The ice fully melts before the water temperature drops below \(0^{\circ} \mathrm{C}\), which is physically feasible. The calculated temperature follows the entire process of melting and heating of the ice, ensuring the final temperature is above \(0^{\circ} \mathrm{C}\).

Key Concepts

Heat Transfer

Heat transfer occurs when there is a difference in temperature between two objects or substances. In the context of calorimetry, which is the study of measuring heat transfer, the heat moves from the hot object to the cold one until thermal equilibrium is reached. In our example, heat initially flows from the hot water, which is at 85°C, to the ice, which is at -18°C. This exchange aims to equalize the temperature of both substances.
The fundamental rule governing this process is that the total heat lost by a warmer object equals the total heat gained by a cooler object. This balance can be expressed as:

• Heat lost by hot water = heat gained by cold ice

This formula aids in predicting the final temperature of a mixed system like our water-ice setup, through solving equations that account for the respective specific heat capacities and the required energy changes.

Specific Heat Capacity

Specific heat capacity is a measure of the amount of heat needed to change the temperature of 1 gram of a substance by 1 degree Celsius. Different substances require different amounts of energy for a temperature change.
For water, the specific heat capacity is 4.184 J/g°C, meaning it takes 4.184 joules to raise 1 gram of water by 1°C. In contrast, ice has a specific heat capacity of 2.03 J/g°C, indicating that less energy is needed to increase its temperature.
Understanding specific heat capacity is crucial for calculating the total energy transfer in our scenario, as it helps determine how much energy the ice absorbs as its temperature rises to the melting point and how much energy the water loses as it cools.

Enthalpy of Fusion

The enthalpy of fusion refers to the amount of energy required to change a substance from a solid to a liquid at constant pressure. This is an essential concept when dealing with phase changes, especially from ice to water.
In our case, the enthalpy of fusion for water is 6.01 kJ/mol. This means that to convert 1 mole of ice at 0°C into water at the same temperature, 6.01 kJ of heat energy is required. Given our ice cube is 25 grams, we calculate the number of moles and then determine the energy required for it to fully melt, using the formula:

• Heat required for melting = moles of ice × 6010 J/mol

This energy contribution must be included in the energy balance calculations to find the system's final temperature after the ice melts.

Phase Change

A phase change occurs when a substance transitions from one physical state to another, such as from solid to liquid. It involves energy exchanges without changing temperature. For ice turning into liquid water, this process is crucial.
During a phase change like melting, energy is absorbed by the substance to break bonds, leading to a state transition rather than a temperature increase. In calorimetry problems, it’s critical to account for phase changes as these significantly impact the energy balance in the system.
The ice must first absorb enough energy to reach 0°C, then additional energy to switch phases from solid to liquid, and finally may continue absorbing heat as liquid until equilibrium is established. This understanding ties together with ensuring calculations are inclusive of phase changes to accurately predict outcomes.