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

So what you would do is go through and add 3 teaspoons(tsp) 6 times because of the 6 tablespoons(tbsp). which will give you an answer of 18. another way is to multiply 3x6=18.