Question

4. Calculate the amount of heat required to convert 155 g of ice at -10oC to steam at 110oC? (Hint: Hfusion = 80cal/g and Hvaporization = 540cal/g - show your work on the left, final answer on the right)

Answer 1

Energy required to raise ice from -10 to 0 degrees Celsius:

Q = m Cp Δt

Q = 155 g x 2.03 J/g C x 10 C

Q1 = 3146.5 J

Energy required to completely melt the ice:

Heat of fusion for water: 333.55 kJ / kg.

Q2 = ΔHf x m

Q2= 333.55 kJ / kg x 155 x 10^-3 kg

Q2 = 51700 J

Energy required to raise water from 0 to 100 Celsius:

Q3 = 155 g x 4.184 J/g*C x 100 C

Q3 = 64852 J

Energy needed to completely vaporize the water:

Heat of vaporization for water: 2257 kJ / kg.

Q4 = ΔHvap x m

Q4 = 2257 KJ / kg x 155 x 10^-3 kg

Q4 = 349835 J

Energy needed to raise steam from 100 to 110 degrees Celsius:

Q5 = 155 g x 1.99 J/g*C x 10

Q5 = 3084.5 J

Now add all the energies:

Qtotal = Q1 + Q2 + Q3 + Q4 + Q5

heat = 3146.5 J + 51700 J + 64852 J + 349835 J + 3084.5 J

heat    = 472618 J

           = 472.6 kJ

           = 113 Kcal