Question

On a trip, you notice that a 3.66 kg bag of ice lasts an average of...

On a trip, you notice that a 3.66 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 -4.79ºC and completely melts to 3.08ºC water in exactly one day? The specific heat of water is 4186 J/kgºC. Specific heat of ice is 2090 J/kgºC and water's heat of vaporization is 334 kJ/kg.

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Answer #1

  energy to heat the ice to melting = mass of ice * specific heat of ice * temperature change
energy to heat the ice to melting = 3.66*2090* 1.71 = 13080.474

energy to melt the ice = mass of ice * heat of fusion for ice/water
energy to melt the ice = 3.66* 334 =1222.44
energy to heat the water to 3.008degrees = mass of water * specific heat of water * temperature change
energy to heat the water to 3.008degrees = 3.66 * 4186 * 1.71 = 26198.4996
Add the three numbers together. Divide by 24 hours (since you are told this takes exactly 1 day). Convert to J/sec. 1 J/sec = 1 W.

avg.power = 1687.55j/sec

avg.power = 1687.55 W

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