An ideal refrigerator utilizes a Carnot cycle operating between 0℃ and 25℃. To turn 10 kg of liquid water at 0℃ into 10 kg of ice at 0℃, (a) how much heat is exhausted into the room and (b) how much energy must be supplied to the refrigerator?

Verified step by step guidance

1. Step 1: Understand the Carnot cycle and its efficiency. The efficiency of a Carnot refrigerator is given by \( \eta = \frac{T_{cold}}{T_{hot} - T_{cold}} \), where \( T_{cold} \) and \( T_{hot} \) are the absolute temperatures (in Kelvin) of the cold and hot reservoirs, respectively. Convert the given temperatures from Celsius to Kelvin: \( T_{cold} = 0 + 273 = 273 \, \text{K} \) and \( T_{hot} = 25 + 273 = 298 \, \text{K} \).
2. Step 2: Calculate the heat required to freeze the water. The latent heat of fusion for water is \( L_f = 334 \, \text{kJ/kg} \). For 10 kg of water, the heat removed from the water to freeze it is \( Q_{cold} = m \cdot L_f \), where \( m \) is the mass of the water. Substitute \( m = 10 \, \text{kg} \) and \( L_f = 334 \, \text{kJ/kg} \) into the formula.
3. Step 3: Relate the work input to the heat removed and heat exhausted using the Carnot cycle. The work input \( W \) is related to the heat removed \( Q_{cold} \) and the heat exhausted \( Q_{hot} \) by the equation \( W = Q_{hot} - Q_{cold} \). Additionally, the Carnot efficiency equation \( \eta = \frac{Q_{cold}}{W} \) can be rearranged to find \( W \).
4. Step 4: Calculate the heat exhausted into the room. Using the relationship \( Q_{hot} = Q_{cold} + W \), substitute the values of \( Q_{cold} \) and \( W \) obtained from the previous steps to find \( Q_{hot} \).
5. Step 5: Summarize the process. The heat exhausted into the room \( Q_{hot} \) and the energy supplied to the refrigerator \( W \) are determined using the principles of thermodynamics and the Carnot cycle. Ensure all units are consistent and verify calculations step by step.

Key Concepts

Carnot Cycle

The Carnot cycle is a theoretical thermodynamic cycle that provides the maximum possible efficiency for a heat engine or refrigerator operating between two temperature reservoirs. It consists of four reversible processes: two isothermal (constant temperature) and two adiabatic (no heat transfer). Understanding this cycle is crucial for calculating the performance of an ideal refrigerator, as it sets the upper limit on efficiency based on the temperatures of the hot and cold reservoirs.

Latent Heat of Fusion

Latent heat of fusion is the amount of heat energy required to change a substance from solid to liquid or vice versa at a constant temperature. For water, this value is approximately 334 kJ/kg. In the context of the question, it is essential to calculate the energy needed to convert 10 kg of liquid water at 0℃ into ice at the same temperature, as this process involves removing heat from the water.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of the efficiency of a refrigerator or heat pump, defined as the ratio of the heat removed from the cold reservoir to the work input required. For a Carnot refrigerator, the COP can be calculated using the temperatures of the hot and cold reservoirs. This concept is vital for determining how much energy must be supplied to the refrigerator to achieve the desired cooling effect.