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