# Question 05bc9

Mar 28, 2017

#### Explanation:

A typical heating curve of water is shown below.

There are five separate heats involved in this problem:

• ${q}_{1}$ = heat required to warm the ice from -15 °C to 0 °C
• ${q}_{2}$ = heat required to melt the ice to water at 0 °C
• ${q}_{3}$ = heat required to warm the water from 0 °C to 100 °C
• ${q}_{4}$ = heat required to boil the water to ice at 1000 °C
• ${q}_{5}$ = heat required to heat the steam from 100 °C to -105.0 °C

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

= mc_1ΔT_1 + mΔ_text(fus)H + mc_3ΔT_3 + mΔ_text(vap)H + mc_5ΔT_5

where

${q}_{1} , {q}_{2} , {q}_{3} , {q}_{4} ,$ and ${q}_{5}$ are the heats involved in each step

$m$ is the mass of the sample

ΔT = T_"f" -T_"i"

${c}_{1} \textcolor{w h i t e}{m m} = \text{the specific heat capacity of ice" = "0.50 cal·°C"^"-1""g"^"-1}$

${c}_{3} \textcolor{w h i t e}{m m} = \text{the specific heat capacity of water" = "1.00 cal·°C"^"-1""g"^"-1}$

${c}_{5} \textcolor{w h i t e}{m m} = \text{the specific heat capacity of steam" = "0.48 cal·°C"^"-1""g"^"-1}$

Δ_text(fus)H = "the enthalpy of fusion of ice" = "79.5 cal·g"^"-1"

Δ_text(vap)H = "the enthalpy of vaporization of water" = "539 cal·g"^"-1"

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

ΔT_1 = "0 °C - (-15 °C)" = "15 °C"

q_1 = mc_1ΔT_1 = 125.00 color(red)(cancel(color(black)("g"))) × 0.50 color(white)(l)"cal"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 15 color(red)(cancel(color(black)("°C"))) = "938 cal"

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

q_2 = 125.00 color(red)(cancel(color(black)("g"))) × 79.5color(white)(l) "cal"·color(red)(cancel(color(black)("g"^"-1"))) = "9938 cal"

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

ΔT = "100 °C - 0 °C" = "100 °C"

q_3 = mcΔT = 125.00 color(red)(cancel(color(black)("g"))) × 1.00 color(white)(l)"cal"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 100 color(red)(cancel(color(black)("°C"))) = "12 500 cal"

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

q_4 = 125.00 color(red)(cancel(color(black)("g"))) × 539color(white)(l) "cal"·color(red)(cancel(color(black)("g"^"-1"))) = "67 380 cal"

${\boldsymbol{q}}_{5}$

ΔT_5 = "105.0 °C - 1000 °C" = "5.0 °C"

q_5 = mcΔT = 125.00 color(red)(cancel(color(black)("g"))) × 0.48 color(white)(l)"cal"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 5.0 color(red)(cancel(color(black)("°C"))) = "300 cal"#

$q = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5} = \text{938 cal" + "9938 cal" + "12 500 cal" + "67 380 cal" + "300 cal" = "91 000 cal" = "91.0 kcal}$

The process releases 91.0 kcal of heat.