Question

A freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80 kg of water at 25.0\(^\circ\)C to 1.80 kg of ice at -5.0\(^\circ\)C in one hour. (a) What amount of heat must be removed from the water at 25.0\(^\circ\)C to convert it to ice at -5.0\(^\circ\)C? (b) How much electrical energy is consumed by the freezer during this hour? (c) How much wasted heat is delivered to the room in which the freezer sits?

Short answer

(a) 393,858 J (b) 164,108 J (c) 557,966 J

Step-by-step solution

Calculate Heat to Cool Water to 0°C

To start, calculate the amount of heat required to lower the temperature of 1.80 kg of water from 25.0°C to 0°C using the specific heat capacity of water, which is 4186 J/kg°C:\[ q_1 = m c \Delta T = 1.80 \times 4186 \times (0 - 25.0) \]\[ q_1 = -1.80 \times 4186 \times 25.0 = -188,370 \text{ J} \]

Calculate Heat to Freeze Water at 0°C

Next, calculate the heat required to freeze the water at 0°C. This requires the latent heat of fusion for water, 334,000 J/kg:\[ q_2 = m L_f = 1.80 \times 334,000 \]\[ q_2 = 601,200 \text{ J} \]

Calculate Heat to Cool Ice to -5°C

Then, calculate the heat required to cool the ice from 0°C to -5.0°C using the specific heat capacity of ice, 2108 J/kg°C:\[ q_3 = m c \Delta T = 1.80 \times 2108 \times (-5.0) \]\[ q_3 = -18,972 \text{ J} \]

Summing Total Heat Removed

The total amount of heat \( Q \) removed is the sum of all three calculated values:\[ Q = q_1 + q_2 + q_3 = -188,370 + 601,200 + (-18,972) \]\[ Q = 393,858 \text{ J} \]

Calculate Electrical Energy Consumed

The coefficient of performance (COP) is given as 2.40. The equation for COP in terms of heat removed \( Q \) and work \( W \) is:\[ \text{COP} = \frac{Q}{W} \Longrightarrow W = \frac{Q}{\text{COP}} \]\[ W = \frac{393,858}{2.40} \]\[ W \approx 164,108 \text{ J} \]

Calculate Wasted Heat Delivered

The wasted heat \( Q' \) delivered to the room is the sum of the work done and the heat extracted:\[ Q' = Q + W = 393,858 + 164,108 \]\[ Q' = 557,966 \text{ J} \]

Key concepts

Heat Transfer

In this context, heat transfer refers to the process of energy moving from one substance to another. Here, it is crucial in changing the water from a liquid at 25°C to ice at -5°C. This process involves three main phases: cooling the water to 0°C, freezing it at 0°C, and then cooling the ice further to -5°C.
Each stage involves different calculations and understanding of concepts. The heat transfer amount depends on both specific heat capacity and latent heat. Understanding how heat is exchanged in these processes helps us calculate the overall energy needed to transform the water.

• Cooling the Water: This phase uses the specific heat capacity of water to compute the heat extracted as water cools to 0°C.
• Freezing the Water: Latent heat is employed here, marking the energy removal needed at constant temperature to change the state from liquid to solid.
• Cooling the Ice: The specific heat of ice is used to determine heat extraction as the temperature of the newly formed ice decreases.

Coefficient of Performance

The coefficient of performance (COP) is a measure of efficiency for refrigerators and freezers. It is defined as the ratio of the heat removed from the freezer to the work input.
In this problem, the freezer's COP is given as 2.40. A higher COP indicates a more efficient system. According to the formula for COP, we identify how much electrical work is done compared to the heat removed.

• Formula Understanding: \[ \text{COP} = \frac{Q}{W} \]Where:
  • \( Q \) is the value of total heat removed (393,858 J).
  • \( W \) is the electrical energy consumed.
• Practical Implications: The calculation helps determine the work necessary to achieve the cooling process, ensuring energy effectiveness.

Latent Heat

Latent heat refers to the energy absorbed or released during a phase change of a substance, without a change in temperature.
In this exercise, latent heat plays a crucial role in transitioning water at 0°C to ice at 0°C. The latent heat of fusion for water is 334,000 J/kg.

• Role in Phase Change: This latent heat must be removed entirely for the liquid water to transform into solid ice.
• Importance: Calculating latent heat accurately ensures you understand the energy exchange in a phase change, which does not depend on the initial or final temperature, but rather the mass of the substance alone.

The energy value is critical for accurate transition estimation from one phase to another, especially in thermal systems like freezers.

Specific Heat Capacity

Specific heat capacity is the amount of heat energy required to raise the temperature of a unit of mass of a substance by one degree Celsius.
In practical terms, it tells us how much energy is needed per kilogram to achieve a temperature increase or decrease.

• Application in this Exercise: Both water and ice have specific heat capacities that determine the energy removal amount required to cool them.
  • Water has a specific heat capacity of 4186 J/kg°C.
  • For ice, it is 2108 J/kg°C.
• Temperature Change Calculation: The specific heat capacity allows calculating the required energy to adjust the temperature of substances, influencing productivity and efficiency in freezing processes.

This concept is essential in defining how substances absorb and release heat energy in transitional temperature intervals, thus playing a crucial role in thermal management systems.