How many joules of heat are needed to change 50.0 grams of ice at 0 C to boil at 120.0 C? Show work

This can be done in the following way;

1 determining the heat required to convert 0° C ice to 0°C water

Heat of fusion of water = 334 J/g

Therefore; Heat = 50 g * 334 J/g = 16700 J

2. Determining the heat required to raise the temperature of water from 0° C to 100°C.

Specific heat of water is 4.18 J/g°C

Change in temperature is 100°C

Therefore; Heat = 50 g * 4.18 J/g°C * 100 = 20900 J

3. Determining the heat required to convert 100 ° C water to 100°C vapor

Heat of vaporization of water = 2257 J/g

Heat = Mass of water * heat of vaporization

Heat = 50 g * 2257 = 112850 J

4. Determining the heat required to go from 100° C to 120° vapor

specific heat of vapor = 2.09 J/g°C

Heat = mass * Specific heat of vapor * change in temperature

= 50 g * 2.09 * (120-100) = 2090 J

Therefore the total heat required is

= 16700 J + 20900 J + 112850 J + 2090 J = 152540 J or 152.54 kJ