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