Problem 66 - General Chemistry

Question

When ice at \(0^{\circ} \mathrm{C}\) melts to liquid water at \(0^{\circ} \mathrm{C}\), it absorbs \(0.334 \mathrm{~kJ}\) of heat per gram. Suppose the heat needed to melt \(31.5 \mathrm{~g}\) of ice is absorbed from the water contained in a glass. If this water has a mass of \(0.210 \mathrm{~kg}\) and a temperature of \(21.0^{\circ} \mathrm{C}\), what is the final temperature of the water? (Note that you will also have \(31.5 \mathrm{~g}\) of water at \(0^{\circ} \mathrm{C}\) from the ice.)

Step-by-Step Solution

Answer

The final temperature of the water is approximately 9.0°C.

Step 1: Understand the Problem

To find the final temperature of the water, we need to calculate how much the water's temperature decreases due to the energy required to melt the ice.

Step 2: Calculate the Heat Required for Melting Ice

The heat required to melt the ice can be calculated using the formula: \( Q = m \times L_f \), where \( Q \) is the heat absorbed, \( m \) is the mass of ice, and \( L_f \) is the latent heat of fusion with a value of \( 0.334 \text{ kJ/g} \). With \( m = 31.5 \text{ g} \), calculate \( Q \): \( Q = 31.5 \text{ g} \times 0.334 \text{ kJ/g} = 10.521 \text{ kJ} \).

Step 3: Convert Heat to Joules

Convert the heat absorbed into joules since SI units will be used for subsequent calculations. Remember the conversion: \( 1 \text{ kJ} = 1000 \text{ J} \). Therefore, \( Q = 10.521 \text{ kJ} \times 1000 = 10521 \text{ J} \).

Step 4: Determine the Water's Mass in Grams

Convert the mass of the water initially in the glass from kilograms to grams: \( 0.210 \text{ kg} = 210 \text{ g} \).

Step 5: Calculate Temperature Decrease of Initial Water

Use the formula: \( Q = m \times c \times \Delta T \), where \( c \) is the specific heat capacity of water \( 4.18 \text{ J/g°C} \), and \( \Delta T \) is the change in temperature. Rearrange to solve for \( \Delta T \): \( \Delta T = \frac{Q}{m \times c} \).

Step 6: Plug in Values for Temperature Decrease

Substitute the known values into the equation to find the temperature decrease: \( \Delta T = \frac{10521 \text{ J}}{210 \text{ g} \times 4.18 \text{ J/g°C}} \approx 12.0 \text{ °C} \).

Step 7: Calculate Final Temperature of the Water

Subtract the temperature decrease from the initial temperature of the water: \( T_{\text{final}} = 21.0 \text{ °C} - 12.0 \text{ °C} = 9.0 \text{ °C} \).

Key Concepts

Understanding Specific Heat Capacity

Specific heat capacity is an important concept in thermodynamics. It helps us understand how substances absorb and release heat. The specific heat capacity of a substance is the amount of heat required to change the temperature of 1 gram of the substance by 1°C. For water, which we frequently deal with, the specific heat capacity is 4.18 J/g°C. This high value means water can absorb a lot of heat without a large change in temperature.

This property is essential when calculating energy transfer in processes like melting ice. By knowing the specific heat capacity, we can calculate how much energy is needed to raise or lower the temperature of a certain mass of water. This forms the basis of many heat transfer calculations.

The Impact of Temperature Change

Temperature change is a key concept when considering how heat affects substance. When heat is added or removed, substances undergo a change in temperature unless they are in a phase transition.

In our scenario where 0.210 kg of water is present at 21.0°C, using the specific heat capacity, we calculate how much the water's temperature drops because of melting ice. This drop is given by the formula \( \Delta T = \frac{Q}{m \times c} \). The subtraction of this temperature change from the initial temperature gives the final temperature.

Understanding how heat affects the temperature helps us predict and control conditions in practical applications like cooking or climate control.

Efficient Heat Transfer Calculation

Heat transfer calculations are essential for understanding energy exchanges in physical systems. Calculating the heat required or released during a process involves understanding mass, specific heat capacity, and temperature change.

In our example, heat transfer calculations help determine how much energy the water loses when melting the ice, backed by specific heat capacity and latent heat of fusion. By rearranging the formula \( Q = m \times c \times \Delta T \), we determine the impact of the heat transferred to the melting ice on the water’s temperature.

• The mass of the water must be correctly measured in grams to match the units.
• The calculation includes conversion between different heat units (e.g., converting kilojoules to joules).

Utilizing these calculations becomes crucial in fields like engineering, meteorology, and environmental science.