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 × #(18.02" g H₂O")/(1" mol H₂O")# = 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"#
Adding them all up
#q = q_1 + q_2 + q_3 + q_4 + q_5 = "(-1.63 – 40.65 – 7.53 – 6.02 – 1.88) kJ" = "-57.71 kJ"#
