# Question 64f37

##### 1 Answer
Sep 5, 2016

The amount of heat required is 12.68 kJ.

#### Explanation:

There are four separate heats involved in this problem:

1. ${q}_{1}$ = heat required to warm the ice from -11.09 °C to 0 °C
2. ${q}_{2}$ = heat required to melt the ice to water at 0 °C
3. ${q}_{3}$ = heat required to warm the water from 0 °C to 100 °C
4. ${q}_{4}$ = 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

${\boldsymbol{q}}_{1}$

$n = \text{0.232 mol}$
${c}_{1} = \text{37.6 J·°C"^"-1""mol"^"-1}$
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"

${\boldsymbol{q}}_{2}$

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"

${\boldsymbol{q}}_{3}$

${c}_{2} = \text{75.2 J·°C"^"-1""mol""-1}$
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"

${\boldsymbol{q}}_{4}$

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"#

$q = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} = \text{0.096 74 kJ" + "1.397 kJ" + "1.745 kJ" + "9.442 kJ" = "12.68 kJ}$