# Question 42ee6

May 21, 2014

Converting 1.00 mol of steam at 145 °C to ice at -50.0 °C involves the loss of
57.71 kJ of energy.

There are five heats to consider:

${q}_{1}$ = heat lost on cooling steam from 145 °C to 100 °C.

${q}_{2}$ = heat lost on condensing steam to water at 100 °C.

${q}_{3}$ = heat lost on cooling water from 100 °C to 0°C.

${q}_{4}$ = heat lost on freezing water to ice at 0 °C.

${q}_{5}$ = heat lost on cooling ice from 0 °C to -50.0 °.

The total heat evolved is

$q = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5}$

1. Cooling the Steam

$m$ = 1.00 mol H₂O × $\left(18.02 \text{ g H₂O")/(1" mol H₂O}\right)$ = 18.02 g H₂O

ΔT = T_2 – T_1 = (100-145) °C = -45 °C

q_1 = mcΔT = 18.02 g × 2.01 J·g⁻¹°C⁻¹ × (-45 °C) = -1629 J = -1.63 kJ

2. Condensing the Steam

Δ H_"cond" = -ΔH_"vap" = "-2256 J·g"^-1

If this problem had come from your text, you would find the heat of vaporization there. For an on-line problem like this, you would look up the value on-line. You can find a table of heats of vaporization here.

q_2 = m Δ H_"cond" = "18.02 g" × ("-2256 J·g"^-1) = "-40 653 J" = "-40.65 kJ"

3. Cooling the Water

ΔT = T_2 – T_1 = "(0 - 100) °C" = "-100 °C"

q_3 = mcΔT = "18.02 g" × "4.18 J·g"^-1"°C"^-1 × "(-100 °C)" = "-7532 J" = "-7.53 kJ"

4. Freezing the Water

Heat of freezing = -Heat of fusion

ΔH_"freeze" = -ΔH_"fus" = "-334 J·g"^-1

q_4 = m Δ H_"freeze" = "18.02 g" × "(-334 J·g"^-1")"= "-6019 J" = "-6.02 k"

5. Cooling the Ice

ΔT = T_2 – T_1 = "(-50.0 - 0) °C" = "-50.0 °C"

q_5 = mcΔT = "18.02 g" × "2.09 J·g"^-1"°C"^-1 × "(-50.0 °C)" = "-1883 J" = "-1.88 kJ"#

$q = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5} = \text{(-1.63 – 40.65 – 7.53 – 6.02 – 1.88) kJ" = "-57.71 kJ}$