how would i calculate the calories used to melt ice and the heat of fusion?

Expt 2- Energy and Matter- Data SIDE Partners in Changes of State CL Mass of empty calorimeter cup with lid Mess of lorimeter with lid water oc C3 Final de temperature 107.15 1 22 135.64 C Initial water temperature 8 C4. Mass of calorimeter with lid + water + melted ice 107.15 -6.835 Calculations: CS. Mass of water 1-22 c C6. Temperature change (AT) 100.315, -19 - 1905.985 - 541.31 C7. Calories lost by water Calories used to melt ice 28.49 64-62 CR Mass of ice that melted C9. Heat of fusion (calories to melt i gofice) 1 calig Show calculations for C7 Calories lost by water, C8 Mass of ice that melted and C9 Heat of fusion: - m. s.45 Cqq head qheat gairogice mass ofice Toss by H20 7- 100.315g. I cal. 19 94 9 9= 28.490 kal. 196 heat of Curion 8 ca 9 where m= mass (@), s= specific heat (2) or) and AT-temperature change (T2-Ti) in C. The specific heat for water is 4.184(e) or 1.00 Heat of fusion is the amount of energy needed to melt a given mass of a solid at its melting point temperature. In the case of ice that melts at a constant temperature of 0 C, the energy needed to melt 1 gram of ice (0 C) is 80 calories. Thus, heat of fusion of ice is 80 cal/g. A same amount of heat would release when liquid water freezes to solid ice at 0 C. Melting (0C): H2O (s) + heat of fusion (80 cal/g) H20 (1) Freezing (0C) H20 (1) H20 (s) + heat of fusion (80 cal/g) In this experiment, ice will be added to water at room temperature. When this occurs, the temperature of the liquid water will drop, indicates the loss of heat by water. This loss of heat by liquid water will be used to melt the ice. - Heat (cal) lost by water (negative value) Heat (cal) gained to melt ice = mass of water x 1.00 cal g x If the amount (mass) of ice is known, the heat of fusion can be determined: Heat of fusion (cal/g) = heat (cal)gained to melt ice grams of ice