Question

What is the total energy (in Joules) required to raise the temperature of 630 g of water from 20degC to 110degC? Please draw this change on a heating curve making sure to label the starting temperature, ending temperature and phase change.

ΔHfus = +6.01kJ/mol
ΔHvap = +40.7 kJ/mol

csteam = 2.02 J/g degC
Cice = 2.01 J/g degC

Answer 1

The question can be dived into 3 parts:

1st part: 20 deg C water to 100deg C water

2nd part: 100 deg C water to 100 deg C steam

3rd part: 100 deg C steam to 110 deg C steam

The formula for energy: E=mcΔT

where m = mass

c = specific heat

ΔT = Change in temperature (T2 - T1)

1st PART:

Taking: c_water = 4.2 J/g/deg C

m = 630g (given)

ΔT = 100 - 20 = 80 deg C

E₁ = 630*4.2*80 = 227.392 kJ

2nd PART:

E₂ = Heat of vapourization x moles

moles of water = 630/18 = 35 moles

E₂ = 40.7 * 35 = 1,424.5 kJ

3rd PART:

Given, c_ice = 2.01 J/g/deg C

ΔT = 110 - 100 = 10 deg C

E₃ = 630*2.01*10 = 12.663 kJ

So, total energy = E₁ + E₂ + E₃ = 1665.095 kilo Joules