How much heat energy, in kilojoules, is required to convert 63.0 g of ice at - 18.0 ∘c to water at 25.0 ∘c?

Mass of ice m = 63 gms

Temperature of Ice T ice = - 18C

Temperature of Water T water = 25C

Specific heat of Ice C ice = 2.09 J/g C

Specific heat of water C water = 4.184 J/g C

Delta H fusion for water = 334 J/g

Now there are three stages

1. Heat energy while converting - 18C Ice to 0C Ice

Q1 = m x C ice x delta T = 63 x 2.09 x (0 - (-18)) = 2370.06 J

2. Heat energy while converting 0C ice to 0C water

Q2 = m x Delta H fusion for water = 63 x 334 = 21042 J

3. Heat energy while converting 0C water to 25C water

Q3 = m x C water x delta T = 63 x 4.184 x (25 - 0) = 6589.8 J

Q = Q1 + Q2 + Q3 = 30001.86 = 30 KJ

Heat Energy required = 30 KJ