91Ó°ÊÓ

Thermodynamic cycles play a crucial role in refrigeration and many other processes. A thermodynamic cycle involves a series of processes that return a system to its initial state, transferring energy in the form of heat and work. For refrigeration cycles, the primary objective is to move heat from a low-temperature region to a high-temperature region. This process requires work input since heat naturally flows from hot to cold regions. Understanding these cycles helps us optimize energy usage and design more effective refrigeration systems. An example is the cycle used in the problem, where heat is transferred from the soft drink to the surroundings.

Heat transfer is the movement of thermal energy from one object or material to another due to a temperature difference. In the problem, heat transfer occurs when the soft drink cools down. The heat removed from the soft drink is quantified using the formula: \[ Q = mcΔT \]

• \( Q \) is the amount of heat removed.
• \( m \) is the mass of the soft drink.
• \( c \) is the specific heat capacity, which is a property of the material.
• \( ΔT \) is the temperature change.

In this example, the heat removed from the soft drink is 23.874 kJ, which is then used to calculate the work input for the refrigeration process.

Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree. It is a critical property in calculating heat transfer. For water, the specific heat capacity (\( c \)) is 4.18 kJ/kg·K. This value indicates that to change the temperature of 1 kg of water by 1 degree Celsius (or Kelvin), 4.18 kJ of energy is needed. In the given problem, the specific heat capacity of water is used to determine the amount of energy transferred to cool the soft drink from 25°C to 6°C. By knowing the mass (0.3 kg) and the temperature change (19°C), we can apply the formula to calculate the energy removed.

The Coefficient of Performance (COP) is a measure of efficiency for refrigeration cycles. It indicates how effectively a refrigeration cycle uses work to transfer heat. The formula for calculating COP in a refrigeration cycle operating between two temperatures is: \[ \text{COP}_{\text{ref}} = \frac{T_L}{T_H - T_L} \] where:

• \( T_L \) is the lower temperature (in Kelvin).
• \( T_H \) is the higher temperature (in Kelvin).

The higher the COP, the more efficient the refrigeration cycle. In the problem, the COP is determined to be 14.684, which is then used to calculate the minimum theoretical work input required to cool the soft drink. The formula is: \[ W = \frac{Q}{\text{COP}} \] Using the heat removed (23.874 kJ) and the COP, we find that the minimum theoretical work input is 1.626 kJ.