On a trip, you notice that a 3.64 kg 3.64 kg bag of ice lasts an average of 1 day 1 day in your cooler. What is the average power absorbed by the ice if the ice starts at 0 °C 0 °C and completely melts to liquid water at 0 °C 0 °C in exactly 1 day 1 day? Give your answer in units of watts.

14.07W

Step-by-step explanation:

Mass of ice = 3.64 kg

Δt = 1 day

Initial temperature = 0°c

Final temperature = 0°c

Convert 1 day to seconds

1 * 24 * 60 * 60 = 8.64 * 10^ 4 secs

Latent heat of fusion of water (Lf) = 334 * 10^3 J/kg

Average power = W/Δt

Energy that leaves and enters the system is heat energy. Therefore,

Average power = Q/Δt

Q = m (ice) * Lf

Q = 3.64 * 334 * 10^3

Power = (3.64 * 334 * 10^3) / 8.64 * 10^4

= 1.21576 * 10^6 / 8.64 * 10^4

= 0.1407 * 10^2

= 14.07 W