Question

A reversible refrigerator has a coefficient of performance of \(3.0 .\) How much work must be done to freeze \(1.0 \mathrm{kg}\) of liquid water initially at \(0^{\circ} \mathrm{C} ?\)

Short answer

Answer: The work required is approximately 111,333.33 Joules.

Step-by-step solution

Calculate energy required to change the water to ice

To determine the required work, we first need to find the amount of energy that the refrigerator has to extract from the water to turn it into ice. This can be calculated using the formula for latent heat of fusion. For water, the latent heat of fusion (Lf) is approximately 334,000 J/kg. Therefore, the energy required to turn 1 kg of water into ice can be given by: Q_cold = mass * Lf Q_cold = (1 kg) * (334,000 J/kg) = 334,000 J

Determine the work done by the refrigerator

Now, we have the energy that needs to be extracted from the water. We can use the formula for the coefficient of performance to determine the work done by the refrigerator: Coefficient_of_performance = Q_cold / W Rearranging, we get: W = Q_cold / Coefficient_of_performance W = 334,000 J / 3.0 W = 111,333.33 J

Final answer

So, the work that must be done to freeze 1 kg of water initially at 0°C using a reversible refrigerator with a coefficient of performance of 3.0 is approximately 111,333.33 Joules.