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On a trip, you notice that a 4.56 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 0°C and completely melts to 0°C water in exactly one day? (1 watt = 1 joule/second or 1 W = 1 J/s.)

Answer :

Answer:

17.63W

Explanation:

The total energy supplied to the ice is used to melt the ice. Since the ice starts to melt at [tex]0^oC[/tex] and also melts completely at the same temperature, the latent heat of fusion will be the focus in solving this problem.

Latent heat of fusion [tex]H_f[/tex] is defined as the amount of energy that changes a solid substance to liquid state without any change in temperature. It is given by the following relationship;

[tex]H_f=ml.....................(1)[/tex]

where m is the mass of the substance and l is the specific latent heat of fusion of the substance.

The specific latent heat of fusion of ice is [tex]334000J/kg[/tex] approximately.

The total heat supplied to the ice was used to melt the ice and is calculated using equation (1), given that m = 4.56kg.

[tex]H_f=4.56*334000\\H_f=1523040J[/tex]

The total energy supplied to the ice is equivalent to the average power P entering the ice multiplied by the time t for which the power is supplied. Therefore;

[tex]P *t=1523040..............(2)[/tex]

where t = 1 day and 1 day = 24 hours.

24 hours = 60 x 60 x 24s =86400s.

Hence, t = 86400s.

Substituting into equation (2), we get the following;

[tex]P*86400=1523040[/tex]

[tex]P=\frac{1523040}{86400}\\P=17.63W[/tex]

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