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On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at 0ºC and completely melts to 0ºC water in exactly one day 1 watt = 1 joule/second (1 W = 1 J/s) ?

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Answer:

[tex]P=13.5 W[/tex]

Explanation:

In this case, power is the rate of transferring heat per unit time:

[tex]P=\frac{Q}{\Delta t}(1)[/tex]

The heat is given by the formula of the latent heat of fusion, since the ice is melting.

[tex]Q=mL_f(2)[/tex]

Here m is the ice's mass and [tex]L_f[/tex] is the heat of fusion of ice. Recall that one day has 86400 seconds. Replacing (2) in (1) and solving:

[tex]P=\frac{mL_f}{\Delta t}\\P=\frac{3.5kg(334*10^3\frac{J}{kg})}{86400s}\\P=13.5 W[/tex]

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