Problem 37 - College Physics

Question

An ice-cube tray contains \(0.350 \mathrm{~kg}\) of water at \(18.0^{\circ} \mathrm{C}\). How much heat must be removed from the water to cool it to \(0.00^{\circ} \mathrm{C}\) and freeze it? Express your answer in joules and in calories.

Step-by-Step Solution

Answer

The heat removed is approximately 90,559.4 J or 21,646 calories.

Step 1: Identify the Mass and Temperature Change

The mass of the water in the ice-cube tray is given as \(0.350 \text{ kg}\). The initial temperature of the water is \(18.0^{\circ} \text{C}\), and it needs to be cooled to \(0.00^{\circ} \text{C}\), indicating a temperature change of \(-18.0^{\circ} \text{C}\).

Step 2: Calculate the Heat Removed to Cool the Water

First, calculate the heat required to cool the water from \(18.0^{\circ} \text{C}\) to \(0.00^{\circ} \text{C}\) using the formula \( Q = mc\Delta T \). Here, \( m = 0.350 \text{ kg} \), \( c = 4186 \text{ J/kg}^{\circ} \text{C} \) (specific heat capacity of water), and \( \Delta T = -18.0^{\circ} \text{C}\). Thus, \( Q = 0.350 \times 4186 \times (-18.0) \).

Step 3: Perform the Calculation

Perform the calculation for the heat removed to cool the water: \[ Q = 0.350 \times 4186 \times (-18.0) = -26340.6 \text{ J} \] The heat is negative since it is removed from the system.

Step 4: Calculate the Heat Removed to Freeze the Water

Next, calculate the heat required to freeze the water at \(0.00^{\circ} \text{C}\) using the formula \( Q = mL_f \). The latent heat of fusion for water \( L_f \) is \(334,000 \text{ J/kg}\). Thus, \( Q = 0.350 \times 334,000 \).

Step 5: Perform the Calculation for Freezing

Calculate the heat removed to freeze the water: \[ Q = 0.350 \times 334,000 = 116,900 \text{ J} \]

Step 6: Total Heat Removed

The total heat removed is the sum of the heat removed during cooling and freezing: \[ Q_{total} = -26340.6 + 116,900 = 90,559.4 \text{ J} \]

Step 7: Convert Joules to Calories

To find the heat in calories, use the conversion factor \( 1 \text{ cal} = 4.184 \text{ J} \). \[ Q_{calories} = \frac{90,559.4}{4.184} \approx 21,646 \text{ cal} \]

Key Concepts

Specific Heat Capacity

Specific heat capacity is a measure of the amount of heat energy needed to change the temperature of a substance. It tells us how much energy, in joules, is required to raise the temperature of one kilogram of a substance by one degree Celsius. This is a unique property for each material, and for water, it is relatively high. This means water requires a lot of energy to change its temperature.
Specific heat capacity is crucial in heat transfer calculations, especially in understanding how substances heat up or cool down. The formula for calculating the change in thermal energy due to temperature change is:\[ Q = mc\Delta T \]where

• \( Q \) is the heat energy transferred (in joules),
• \( m \) is the mass of the substance (in kilograms),
• \( c \) is the specific heat capacity (in joules per kilogram per degree Celsius), and
• \( \Delta T \) is the temperature change (in degrees Celsius).

Latent Heat of Fusion

Latent heat of fusion is the heat energy required to change a substance from a solid to a liquid, or vice versa, without changing its temperature. For water, this phase change occurs at 0°C, where ice melts into water or water freezes into ice. The energy needed for this transformation is called the latent heat of fusion.
This concept is critical as it explains why we need additional energy to change the state of a substance, beyond just changing its temperature. For water, this latent heat is quite high, at 334,000 J/kg.
During this phase change, the temperature of the water remains constant, even though the energy is being added or removed. The formula to calculate the heat involved in this phase change is:\[ Q = mL_f \]where

• \( Q \) is the heat energy involved in the phase change (in joules),
• \( m \) is the mass (in kilograms), and
• \( L_f \) is the latent heat of fusion (in joules per kilogram).

Temperature Change

Temperature change refers to the difference in temperature a substance undergoes, typically when it gains or loses thermal energy. This change is directly involved in calculating the amount of energy needed or released as a substance's temperature changes.
When calculating the heat energy exchange during a temperature change, it's important to know:
- The initial and final temperatures of the substance.- The substance mass.- Its specific heat capacity.
The formula incorporates these variables and is calculated using:\[ Q = mc\Delta T \]Where the temperature change \( \Delta T \) is calculated as the final temperature minus the initial temperature.
You can figure out the amount of energy transferred because of this change, using the above formula, which shows us that not only the size of the temperature change but also the material's properties determine how much energy is needed.

Thermal Energy

Thermal energy is the total internal energy contained in a substance, related to its temperature and phase (like whether it is solid, liquid, or gas). It's important to note that thermal energy is the sum of both kinetic energy, due to the motion of particles, and potential energy, due to forces between them.
In our context, removing thermal energy results in cooling and potentially freezing the substance. The total thermal energy change involves both cooling the substance and changing its phase, if applicable.
The complete process of removing heat from the water in an ice-cube tray involves two main components:

• Cooling the water, computed with \( Q = mc\Delta T \).
• Freezing the water, using the phase change formula \( Q = mL_f \).

Recognizing the roles these calculations play in understanding thermal energy helps grasp how energy transfers occur in different circumstances, leading to changes in temperature and state of matter.