Question

Based on the thermodynamic properties provided for water, determine the amount of energy needed for 269 g of water to go from 85.5 °C to 161 °C.

Property	Value	Units
Melting point	0.0	°C
Boiling point	100.0	°C
ΔHfusΔHfus	6.01	kJ/mol
ΔHvapΔHvap	40.67	kJ/mol
cp (s)	37.1	J/mol·°C
cp (l)	75.3	J/mol·°C
cp (g)	33.6	J/mol·°C

Answer 1

Answer – We are given,

mass of water = 269 g ,

initial temp, ti = 85.5 ^oC ,

final temp, tf = 161 ^oC.

ΔHvap = 40.67 kJ/mol,

Boling point = 100.0 ^oC ,

ΔHfus = 6.01 kJ/mol

Specific heat liquid = 75.3 J/mol.°C

Specific heat gas = 33.6 J/mol.°C

We are given values per mole of water, so need to convert the given mass into there mole.

Moles of water = 269 g / 18.016 g.mol^-1

                    = 14.93 mol

First we need to calculate heat from 85.5 °C to 100.0 °C

We know formula

q1 = m*C*Δt

     = 14.93 mol x 75.3 J/mol.°C x (100 -85.5) °C

    = 16302.6 J

    = 16.303 kJ

we need to calculate heat from 100.0 °C to 100.0 °C

We know formula

q2 = m* ΔHvap

     = 14.93 mol x 6.01 kJ/mol

    = 89.74 kJ

we need to calculate heat from 100 °C to 161.0 °C

We know formula

q3 = m*C*Δt

     = 14.93 mol x 33.6 J/mol.°C x (161 -100) °C

    = 30602.9 J

     = 30.603 kJ

So total heat

q = q1 + q2+q3

   = 16.303 kJ + 89.74 kJ + 30.603 kJ

   = 136.6 kJ

Heat needed to raise the temperature of the sample 269 g of water to go from 85.5 °C to 161 °C   is 136.6 kJ