Problem 24 - College Physics

Question

\(\bullet\) A Carnot freezer that runs on electricity removes heat from the freezer compartment, which is at \(-10^{\circ} \mathrm{C}\) , and expels it into the room at \(20^{\circ} \mathrm{C}\) . You put an ice-cube tray con- taining 375 \(\mathrm{g}\) of water at \(18^{\circ} \mathrm{C}\) into the freezer. (a) What is the coefficient of performance of this freezer? (b) How much energy is needed to freeze this water? (c) How much electrical energy must be supplied to the freezer to freeze the water? (d) How much heat does the freezer expel into the room while freezing the ice?

Step-by-Step Solution

Answer

COP is 8.77; 153,447 J needed to freeze water; 17,489 J electrical energy; 170,936 J heat expelled.

Step 1: Calculate the Temperatures in Kelvin

To find the coefficient of performance for a Carnot freezer, we first need to convert the Celsius temperatures to Kelvin. The conversion formula is given by:\[ T_K = T_C + 273.15 \]For the freezer compartment:\[ T_{ ext{cold}} = -10 + 273.15 = 263.15 \, K \]For the room:\[ T_{ ext{hot}} = 20 + 273.15 = 293.15 \, K \]

Step 2: Calculate the Coefficient of Performance (COP)

The coefficient of performance (COP) for a Carnot refrigerator is given by the formula:\[ \text{COP} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}} \]Substitute the temperatures we calculated:\[ \text{COP} = \frac{263.15}{293.15 - 263.15} = \frac{263.15}{30} \approx 8.77 \]

Step 3: Calculate Energy to Freeze Water

To find the energy needed to freeze the water, calculate the energy required to decrease the water temperature to 0°C and then freeze it. First, use the specific heat capacity formula:\[ Q_1 = mc\Delta T \]Where:- \( m = 0.375 \, \text{kg} \)- \( c = 4186 \, \text{J/kg°C} \)- \( \Delta T = 18 \, °C \)Calculate \( Q_1 \):\[ Q_1 = 0.375 \times 4186 \times 18 \approx 28,197 \, \text{J} \]Next, calculate the energy to freeze water at 0°C using the heat of fusion:\[ Q_2 = mL_f \]Where \( L_f = 334,000 \, \text{J/kg} \)Calculate \( Q_2 \):\[ Q_2 = 0.375 \times 334,000 = 125,250 \, \text{J} \]Total energy \( Q \) is:\[ Q = Q_1 + Q_2 = 28,197 + 125,250 = 153,447 \, \text{J} \]

Step 4: Calculate Electrical Energy Needed

Given the COP, the electrical energy \( W \) supplied is related by:\[ \text{COP} = \frac{Q}{W} \]Therefore:\[ W = \frac{Q}{\text{COP}} = \frac{153,447}{8.77} \approx 17,489 \, \text{J} \]

Step 5: Calculate Heat Expelled to the Room

The heat expelled \( Q_h \) to the room is the sum of the heat removed from the freezer compartment \( Q \) and the work done:\[ Q_h = Q + W = 153,447 + 17,489 = 170,936 \, \text{J} \]

Key Concepts

Carnot Cycle

The Carnot cycle represents an idealized thermodynamic cycle that provides the maximum possible efficiency for a heat engine operating between two temperatures. It is comprised of two isothermal processes and two adiabatic processes. In the context of a refrigerator or freezer, such as the one in the exercise, the Carnot cycle explains how heat is absorbed at a low temperature and expelled at a higher temperature, as the system does work to transfer energy. This cycle is essential for understanding why the theoretical efficiency limits are not always reached in real-world applications, helping us to appreciate the significance of the Carnot cycle as a model for improving system efficiencies. Key characteristics of the Carnot cycle involve:

• Absorption of heat at a constant temperature from the cold reservoir.
• Expansion of the gas doing work at constant entropy.
• Expulsion of heat at a constant temperature to the hot reservoir.
• Compression of the gas to its original state, again at constant entropy.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of an appliance's efficiency, particularly in refrigeration cycles. It indicates how effectively a system is using energy to cool an environment, calculated as the ratio of heat removed from the cold space to the work input. For a Carnot refrigerator, this is represented as:\[ \text{COP} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}} \]where:

• \( T_{\text{cold}} \) is the absolute temperature of the cold reservoir (freezer compartment in this case).
• \( T_{\text{hot}} \) is the absolute temperature of the hot reservoir (the room).

In our scenario, understanding COP helps us quantify the performance of a freezer, denoting how much more heat it can remove per unit of electrical energy consumed, aiding in evaluating and optimizing energy use.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or material to another, driven by a temperature difference. In scenarios like a freezer, heat transfer is pivotal because it dictates the energy exchanges necessary for cooling food items. There are three primary modes of heat transfer:

• Conduction, which occurs through direct contact.
• Convection, which involves the movement of fluids or gases.
• Radiation, which utilizes electromagnetic waves.

In this exercise, heat transfer encompasses the movement of heat from the ice cube's water into the freezer's cooling system, effectively reducing the water's temperature until it freezes.

Specific Heat Capacity

Specific heat capacity is a property of a material that describes how much heat energy is required to change the temperature of a unit mass by one degree Celsius. It highlights how different substances require varying amounts of energy to change their temperatures. The formula used for specific heat capacity calculations is:\[ Q = mc\Delta T \]where:

• \( Q \) represents the heat added or removed.
• \( m \) is the mass of the substance.
• \( c \) stands for specific heat capacity (e.g., for water \( 4186 \, \text{J/kg°C} \)).
• \( \Delta T \) is the temperature change.

For water turning into ice in the freezer, applying these calculations helps us determine how much heat energy must be removed to cool the water from its initial temperature to freezing.

Heat of Fusion

The heat of fusion refers to the amount of energy needed to change a substance from a solid to a liquid at its melting point, without a change in temperature. In the exercise, it's crucial for understanding the energy requirements when freezing water. The formula to calculate the energy needed using the heat of fusion is:\[ Q = mL_f \]where:

• \( Q \) is the heat energy involved in the phase change.
• \( m \) is the mass of the water.
• \( L_f \) is the latent heat of fusion (for water, \( 334,000 \, \text{J/kg} \)).

This energy must be extracted from the water to transition it into ice at 0°C, emphasizing the importance of heat of fusion in freezing processes.