Phase change equations

Question

how do I Calculate the amount of energy needed to change 207 g of ice at -10 degree C to steam at 125 degrees C

Answer 1

In order to find the total energy needed, you’re going to have to find FIVE different heat energies, then ADD them all together.

1. Energy required to heat -10˚C of ice to 0˚C of ice 
2. Energy required to convert 0˚C of ice to 0˚C of WATER
3. Energy required to heat 0˚C of water to 100˚C of water
4. Energy required to convert 100˚C of water to 100˚C of STEAM
5. Energy required to heat 100˚C of steam to 125˚C of steam

Equations You Need

1. q = mcΔT (to change temperature)
2. q = m·ΔHf (to change phase)

Where q = heat energy and m = mass of your sample (207g)

Constants You’ll Need

1. heat of fusion of water (Hf) = 334 J/g
2. heat of vaporization of water (Hv) = 2257 J/g
3. specific heat of ice (c) = 2.09 J/g·°C
4. specific heat of water (c) = 4.18 J/g·°C
5. specific heat of steam (c) = 2.09 J/g·°C

Now to plug all your information into the equation.

1. Find the energy required to heat -10˚C of ice to 0˚C of ice
2. q = mcΔT
3. q = (207 g) x (2.09 J/g·°C) x (100°C - 0°C)
4. q = (207 g) x (2.09 J/g·°C) x (100°C)
5. q = 43,263 J 
6. Energy required to convert 0˚C of ice to 0˚C of water (FUSION)
7. q = m·ΔHf
8. q = (207 g) x (334 J/g)
9. q = 69,138 J
10. Energy required to heat 0˚C of water to 100˚C of water
11. q = mcΔT
12. q = (207 g) x (4.18 J/g·°C) x (100°C - 0°C)
13. q = (207 g) x (4.18 J/g·°C) x (100 °C)
14. q = 86,526 J 
15. Energy required to convert 100˚C of water to 100˚C of steam (VAPORIZATION)
16. q = m·ΔHv
17. q = (207 g) x (2257 J/g)
18. q = 467,199 J
19. Energy required to heat 100˚C of steam to 125˚C of steam
20. q = mcΔT
21. q = (207 g) x (2.09 J/g·°C) x (125 °C - 100 °C)
22. q = (207 g) x (2.09 J/g·°C) x (25°C)
23. q = 10,815.75 J
24. Add them up to find the Heat_TOTAL
25. HeatTOTAL = 43,263 J + 69,138 J + 86,526 J + 467,199 J + 10,815.75 J
26. HeatTOTAL = 676,941.75 J or 676.941 kJ