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 powe r 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.
1 answer:
Answer:
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
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