There are four separate heats involved in this problem:
- q_1q1 = heat required to warm the ice from -11.09 °C to 0 °C
- q_2q2 = heat required to melt the ice to water at 0 °C
- q_3q3 = heat required to warm the water from 0 °C to 100 °C
- q_4q4 = heat required to convert the water to steam at 100 °C
q = q_1 + q_2 + q_3 + q_4 = nc_1ΔT_1 +nΔ_text(fus)H + nc_2ΔT_2 + nΔ_text(vap)H
bbq_1
n = "0.232 mol"
c_1 = "37.6 J·°C"^"-1""mol"^"-1"
ΔT = "0.0 °C - (-11.09 °C)" = "11.09 °C"
q_1 = ncΔT = 0.232 color(red)(cancel(color(black)("mol"))) × 37.6 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""mol"^"-1"))) × 11.09 color(red)(cancel(color(black)("°C"))) = "96.74 J" = "0.096 74 kJ"
bbq_2
Δ_"fus"H = "40.7 kJ·mol"^"-1"
q_2 = 0.232 color(red)(cancel(color(black)("mol"))) × 6.02color(white)(l) "kJ"·color(red)(cancel(color(black)("mol"^"-1"))) = "1.397 kJ"
bbq_3
c_2 = "75.2 J·°C"^"-1""mol""-1"
ΔT_2 = "100 °C - 0°C" = "100 °C"
q_3 = ncΔT = 0.232 color(red)(cancel(color(black)(" mol"))) × 75.2 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""mol"^"-1"))) × 100 color(red)(cancel(color(black)("°C"))) = "1745 J" = "1.745 kJ"
bbq_4
Δ_"vap"H = "40.7 J·K"^"-1""mol""-1"
q_4 = nΔ_"vap"H = 0.232 color(red)(cancel(color(black)("mol"))) × 40.7color(white)(l) "kJ"·color(red)(cancel(color(black)("mol"^"-1"))) = "9.442 kJ"
q = q_1 + q_2 + q_3 + q_4 = "0.096 74 kJ" + "1.397 kJ" + "1.745 kJ" + "9.442 kJ" = "12.68 kJ"
