Final answer:
The amount of heat that must be removed from the water to convert it to ice is 789.30 kJ. The electrical energy consumed by the freezer during this process is 1894.32 kJ, and the wasted heat delivered to the room is -1105.02 kJ.
Step-by-step explanation:
To calculate the amount of heat that must be removed from the water at 25°C to convert it to ice at -5°C, we can use the formula Q = mcΔT, where Q is the amount of heat energy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, we need to calculate the amount of heat energy required to cool the water from 25°C to 0°C. Using the specific heat capacity of water (4.18 J/g°C) and the mass of the water (1.80 kg), we get:
Q1 = (1.80 kg)(4.18 J/g°C)(25°C - 0°C) = 188.10 kJ.
Next, we need to calculate the amount of heat energy required to freeze the water at 0°C. The heat of fusion of water is 334 J/g, so:
Q2 = (1.80 kg)(334 J/g) = 601.20 kJ.
To find the total amount of heat energy that must be removed, we add Q1 and Q2:
Q = Q1 + Q2 = 188.10 kJ + 601.20 kJ = 789.30 kJ.
To calculate the electrical energy consumed by the freezer, we can use the formula W = COPQ, where W is the work done, COP is the coefficient of performance, and Q is the amount of heat energy.
W = (2.40)(789.30 kJ) = 1894.32 kJ.
The wasted heat delivered to the room can be calculated by subtracting the electrical energy consumed from the total amount of heat energy:
Qwasted = Q - W = 789.30 kJ - 1894.32 kJ = -1105.02 kJ.
Therefore, the amounts of heat energy that must be removed from the water, electrical energy consumed by the freezer, and wasted heat delivered to the room are Q = 789.30 kJ, W = 1894.32 kJ, and Qwasted = -1105.02 kJ respectively.