Question

What quantity of energy does it take to convert 0.200 kg ice at –20.°C to steam at 250.°C? Specific heat capacities: ice, 2.03 J/g·°C; liquid, 4.2 J/g·°C; steam, 2.0 J/g·°C; = 40.7 kJ/mol;  = 6.02 kJ/mol.

What is the energy in kJ

Answer 1

We need to find the amount of heat:
a. required to raise the temperature of ice from -20°C to 0°C.
b. required to melt ice at 0°C.
c. required to raise the temperature of liquid water from 0°C to 100°C.
d. required to evaporate liquid water at 100°C.
e. required to raise the temperature of steam from 100°C to 250°C.

Solving for a:
Q = cmΔT
Q = (2.03 J/g°C)(200 g)(20°C)
Q = 8120 J

Solving for b:
Q = ΔHfus * mole
Q = (6020 J/mol) * (200 g / 18.02 g/mol)
Q = 66815 J

Solving for c:
Q = cmΔT
Q = (4.2 J/g°C)(200 g)(100°C)
Q = 84000 J

Solving for d:
Q = ΔHvap * mole
Q = (40700 J/mol)(200 g / 18.02 g/mol)
Q = 451720 J

Solving for e:
Q = cmΔT
Q = (2.0 J/g°C)(200 g)(150°C)
Q = 60000 J

add all the steps

8120J + 66815J + 84000 J + 451720 J + 60000J

= 670655 J

= 670.655 KJ