How much heat is added to 10.0 g of ice at -200°C to convert it to steam at 120.00°C?

Four phases make up the standard heating/cooling curve. The specific heat (oC/g) of the compound in each phase—solid, liquid, or gas—and the interphase transition energies determine how much a temperature changes within a single phase.

Therefore, in this instance, we must heat the solid water from -200oC to 0oC, then at 0oC, change its phase from solid to liquid, and then heat the liquid water from 0oC to 100oC. Finally, we must heat the gas phase of the water from 100oC to 120oC.

Heat of Fusion: 333.55 J/g; Heat of Vaporization: 2257 J/g; Solid Specific Heat: 2.05 J/g-oK; Liquid Specific Heat: 4.178 J/g-oK; Vapor Specific Heat: 1.89 J/g-oK

One kilogramme * (200°C) * 2.05 J/g-oC + one kilogramme * 333.55 J/g + one kilogramme * (100°C) * 4.178 J/g-oC + one kilogramme * 2257 J/g + one kilogramme * (20°C) * 1.89 J/g-oC

10.0j * (205 J/g-oC * 200oC * 333.55 J/g + 4.178 J/g-oC * 100oC * 2257 J/g + 20oC * 1.89 J/g-oC))

10.02 grams * (410 grams + 333.55 grams + 417.8 grams + 2257 grams + 37.8 grams)

34561.5 Joules

https://tutor.hix.ai provides a good explanation of this kind of problem.

[ q = m \times \left( \Delta H_{\text{fusion}} + \Delta H_{\text{vaporization}} + \Delta T_{\text{final}} \times c_{\text{water}} \right) ] [ q = 10.0 , \text{g} \times (334 , \text{J/g} + 2260 , \text{J/g} + 120.00 , \text{°C} \times 4.18 , \text{J/g°C}) ]