Final answer:
To equate the heat put by a refrigerator into a kitchen when freezing 1.18 kg of water, the calculations involve finding the energy required to cool and freeze the water and dividing by the coefficient of performance of the refrigerator. The 2.33-kW space heater must then run for approximately 55.36 seconds to emit the equivalent amount of energy.
Step-by-step explanation:
To determine how long a 2.33-kW space heater would have to run to equate the energy put into a kitchen by a refrigerator when freezing 1.18 kg of water from 20.0 °C to ice at 0.0 °C, we must first calculate the amount of energy the refrigerator uses. The specific heat of water is approximately 4.18 kJ/(kg·°C), and the latent heat of fusion for water is about 334 kJ/kg.
We use the following formula to calculate the energy required to cool the water to 0 °C and then freeze it:
Energy = (mass x specific heat x temperature change) + (mass x latent heat of fusion)
For the cooling:
Energy = (1.18 kg x 4.18 kJ/kg·°C x (20.0 °C - 0.0 °C)) = 98.44 kJ
For the freezing:
Energy = (1.18 kg x 334 kJ/kg) = 394.12 kJ
Total energy = 98.44 kJ + 394.12 kJ = 492.56 kJ
Since the refrigerator has a coefficient of performance (COP) of 3.82, the work input needed by the refrigerator is:
Work = Total energy / COP
Work = 492.56 kJ / 3.82 = 128.99 kJ
To find out how long the space heater must run to produce the same amount of energy, we divide the total energy by the heater's power (note that 1 kW = 1 kJ/s):
Time = Energy / Power
Time = 128.99 kJ / 2.33 kW = 55.36 seconds
Hence, the 2.33-kW space heater would need to operate for approximately 55.36 seconds to emit the same amount of energy that the refrigerator does when it freezes 1.18 kg of water at 20.0 °C into ice at 0.0 °C.