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.

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