Determine the amount of heat (in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero degrees Celsius, melts at zero degrees Celsius, and boils at 100 degrees Celsius. Remember that you need to take into account three changes: melting ice, heating water, and vaporizing the water.)

Following are important constant that used in present calculations

Heat of fusion of H2O = 334 J/g

Heat of vaporization of H2O = 2257 J/g

Heat capacity of H2O = 4.18 J/gK

Now, energy required for melting of ICE = 334 X 5.25 = 1753.5 J ... (1)

Energy required for raising the temperature water from 0 oC to 100 oC = 4.18 X 5.25 X 100 = 2195.18 J ... (2)

Lastly, energy required for boiling water = 2257X 5.25 = 11849.25 J ... (3)

Thus, total heat energy required for entire process = (1) + (2) + (3)

= 1753.5 + 2195.18 + 11849.25

= 15797.93 J

= 15.8 kJ

Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.