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When water changes from a liquid to a solid, it undergoes a process called fusion. During this process, heat energy is required to change the state of matter without changing its temperature. This heat energy is known as the latent heat of fusion. For water, this value is crucial because it quantifies the energy needed to turn ice into water and vice versa at 0°C.

The latent heat of fusion for water is 334,000 Joules per kilogram (J/kg). This means you need 334,000 joules of energy to convert one kilogram of ice at its melting point completely into liquid water. In the exercise we are given, 85 kilograms of water are being frozen, indicating that the energy absorbed by the system to freeze the water is:

• Calculate using: \( Q_c = m \times L_f \)
• Substitute values: \( Q_c = 85.0 \, \mathrm{kg} \times 334,000 \, \mathrm{J/kg} \)
• Total heat absorbed: \( Q_c = 28,390,000 \, \mathrm{J} \)

This energy is removed from the water during freezing, and because the temperature doesn't drop during the phase change, the energy accommodates the change in state only.

Thermodynamic efficiency is a measure of how well an engine or cycle converts input heat into useful work or, in a refrigeration cycle, how efficiently it converts the work done into heat transfer from the cold reservoir. In a Carnot cycle, which operates only between two temperatures, this efficiency is governed by the temperature of the heat source and the heat sink.

The formula for the efficiency of a Carnot cycle refrigerator or heat pump is:

• Expression: \( \frac{Q_c}{Q_h} = \frac{T_c}{T_h} \)
• Where \( Q_c \) is the heat absorbed from the cold source, and \( Q_h \) is the heat expelled to the warm source.
• \( T_c \) and \( T_h \) are the absolute temperatures (in Kelvin) of the cold and hot reservoirs, respectively.

In the problem, water is frozen, transferring heat at 0°C, or 273.15 K, while room temperature is 24°C, equivalent to 297.15 K. The efficiency equation helps to establish the relationship between the heat absorbed from the water and the heat expelled to the room, allowing us to calculate how much heat is rejected by finding \( Q_h \). This heat rejection indicates how effective the cycle is at transferring thermal energy.

Calculating heat transfer in thermodynamic systems often requires considering energy inputs and outputs. The Carnot cycle provides a theoretical framework for determining these heat exchanges with maximum possible efficiency.

For the ice-making machine scenario, the heat discharged into a warmer room and the work done by the system are our primary focus. Given the specifics of the Carnot cycle, we link the temperatures of the cold and hot reservoirs to find these values effectively:

• To find \( Q_h \) (heat transferred to the room), use the formula: \( Q_h = Q_c \frac{T_h}{T_c} \)
• Given values: \( Q_c = 28,390,000 \, \mathrm{J} \), \( T_c = 273.15 \, \mathrm{K} \), \( T_h = 297.15 \, \mathrm{K} \).

Substituting:

• \( Q_h = 28,390,000 \, \mathrm{J} \times \frac{297.15}{273.15} = 30,914,983 \, \mathrm{J} \)
• This represents the energy evacuated to the room during the cycle.

Moreover, the work done by the cycle, \( W \), is calculated as the difference \( W = Q_h - Q_c \). The result, 2,524,983 J, represents the energy input required to perform this refrigeration process, showcasing the intricacies of heat management in thermal systems.