Question

Phase changes also involve changes in enthalpy (H). Assume an ideal system: Calculate the total H to bring m=46.884g of ice at T=0 deg C to steam at T=100 deg C. Show your work. (Hint: Cp values vary based on states of matter and energy is required to execute a phase change.)

Answer 1

m=46.884g of ice at T=0 deg C to steam at T=100 deg C

1. specific heat of ice = 2.01 J/g·°C

2. heat of fusion of water = 6.00 kJ/mol

3. specific heat of water = 4.18 J/g·°C

4. heat of vaporization of water = 40.657 kJ/mol

5. specific heat of steam = 2.01 J/g·°C

total Heat change =

    (mass H2O) * (specific heat of solid H2O) * (T2 - T1) +

    (mass H2O) * (heat of fusion for solid H2O) +

    (mass H2O) * (specific heat of liquid H2O) + (T3 - T2) +

    (mass H2O) * (heat of vaporization for liquid H2O) +

    (mass H2O) * (specific heat of gaseous H2O) + (T4 - T3)

ice->water then water->steam,

H₂O (s) => H₂O (l) => H₂O (g)

T1 = 0 deg C

T2 = o deg C

T3 =100 deg C

T4 = 100 deg C

Total Heat change = (46.884g) x (2.01 J/g·°C)x(0-0) +

(46.884g) x(6.00 kJ/mol)(1000 J/kJ) * (1 mol H2O / 18.02 g H2O) +

(46.884g) x(4.18 J/g·°C )x(100-0)+

(46.884g) x (40.657 kJ/mol) (1000 J/kJ) * (1 mol H2O / 18.02 g H2O) +

(46.884g) x(2.01 J/g·°C)x(100-100)

= 0+ 15610.654 + 19597.512+ 105780.3988+0

=140988.5648901 Joules

or 1.40988x 10^5 joules