Show the calculation of the energy involved in melting 120 grams of ice at 0oc if the heat of fusion for water is 0.334 kj/g.

The energy involved in melting 120 g of ice = 40080 J

Explanation:

Heat of fusion of water: Heat of fusion is the heat Energy  required  to convert a substance from solid to its liquid form without a change in energy.

The S.I unit of heat of fusion is J/kg.

Q = lm equation 1

Where Q = Energy involved in melting the ice, l = heat of fusion of water, m =  mass of water.

Given:  l = 0.334 kJ/g, m = 120 g

Substituting these values into equation 1

Q = 0.334×120

Q = 40.08 kJ

Q = 40080 J

Therefore the energy involved in melting 120 g of ice = 40080 J

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