Questions: How long will it take (a) a 1000 W heater to evaporate 1.0 kg of water that is already at 100°C, its normal boiling point (b) a refrigerator to freeze 0.40 kg of water that is already at 0°C, its normal freezing point, if it can remove energy at an average rate of 75 W?

Solution Steps

To evaporate water, we need to use the latent heat of vaporization. The latent heat of vaporization for water is approximately \(2260 \, \text{kJ/kg}\).

For 1.0 kg of water, the energy required is: \[ Q = m \cdot L_v = 1.0 \, \text{kg} \times 2260 \, \text{kJ/kg} = 2260 \, \text{kJ} \]

The power of the heater is given as 1000 W, which is equivalent to 1000 J/s. To find the time \(t\) required, we use the formula: \[ t = \frac{Q}{P} = \frac{2260 \times 10^3 \, \text{J}}{1000 \, \text{J/s}} = 2260 \, \text{s} \]

To freeze water, we need to use the latent heat of fusion. The latent heat of fusion for water is approximately \(334 \, \text{kJ/kg}\).

For 0.40 kg of water, the energy required is: \[ Q = m \cdot L_f = 0.40 \, \text{kg} \times 334 \, \text{kJ/kg} = 133.6 \, \text{kJ} \]

The power of the refrigerator is given as 75 W, which is equivalent to 75 J/s. To find the time \(t\) required, we use the formula: \[ t = \frac{Q}{P} = \frac{133.6 \times 10^3 \, \text{J}}{75 \, \text{J/s}} = 1781.3333 \, \text{s} \]

Final Answer