There are five heats to consider:
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q_1q1 = heat lost on cooling steam from 110.0 °C to 100 °C.
q_2q2 = heat lost on condensing steam to water at 100 °C.
q_3q3 = heat lost on cooling water from 100 °C to 0°C.
q_4q4 = heat lost on freezing water to ice at 0 °C.
q_5q5 = heat lost on cooling ice from 0 °C to -40.0 °C.
The total heat evolved is
q = q_1 + q_2 + q_3 + q_4 + q_5q=q1+q2+q3+q4+q5
1. Cooling the Steam
m = "32.0 g H"_2"O"m=32.0 g H2O
For steam, the specific heat capacity, c = "2.010 J·g"^"-1""°C"^"-1"c=2.010 J⋅g-1°C-1.
ΔT = T_2 – T_1 = "(100.0 - 110.0) °C" = "-10.0 °C"
q_1 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "2.010 J·"color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-10.0" color(red)(cancel(color(black)("°C")))) = "-643 J"
2. Condensing the Steam
"Heat of condensation = -Heat of vaporization"
Δ H_"cond" = ""-ΔH_"vap" = "-2260 J·g"^"-1"
q_2 = m Δ H_"cond" = 32.0 color(red)(cancel(color(black)("g")))×("-2260 J·"color(red)(cancel(color(black)("g"^"-1")))) = "-72 320 J"
3. Cooling the Water
For liquid water, the specific heat capacity, c = "4.184 J·°C"^"-1""g"^"-1".
ΔT = T_2 – T_1 = "(0 - 100) °C" = "-100 °C"
q_3 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "4.184 J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1")))× ("-100"color(red)(cancel(color(black)("°C")))) = "-13 389 J"
4. Freezing the Water
"Heat of freezing = -Heat of fusion"
"-"ΔH_"fus" = "334 J·g"^"-1"
ΔH_"freeze" = "-"ΔH_"fus" = "-334 J·g"^-1"
q_4 = "-"m Δ H_"fus" = 32.0 color(red)(cancel(color(black)("g"))) × "-334 J·"color(red)(cancel(color(black)("g"^"-1"))) = "-10 689 J"
5. Cooling the Ice
The specific heat capacity of ice, c = "2.03 J·°C"^"-1""g"^"-1"
ΔT = T_2 – T_1 = "(-40.0 - 0) °C" = "-40.0 °C"
q_5 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × 2.03 "J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × (color(red)(cancel(color(black)("-40.0 °C")))) = "-2598 J"
Adding them all up
q = q_1 + q_2 + q_3 + q_4 + q_5 = "(-643 – 72 320 – 13 389 – 10 689 - 2598) J" = "-99 600 J"
