Question

4. What is the total mass of ice that can be vaporized by 2100 kJ of heat energy?

Answer 1

Answer:

5 g

Explanation:

The heat required to vaporize ice is the sum of

i) Heat required to melt ice at 0°C

ii) Heat required to raise the temperature from  0°C to 100°C

iii) Heat required to vaporize water at 100°C

Thus;

H = nLfus + ncθ + nLvap

H= n(Lfus + cθ + Lvap)

Lfus = 6.01 kJ/mol

Lvap = 41 kJ/mol

c = 75.38

n =?

2100 = n(6.01  + 75.38(100) + 41)

n = 2100 KJ/7585.01 kJ/mol

n = 0.277 moles

Mass of water = number of moles * molar mass

Mass of water = 0.277 moles * 18 g/mol

Mass of water =  5 g