# Question c4205

Oct 27, 2016

The amount of heat given off is 59 000 J.

#### Explanation:

There are three separate heats involved in this problem:

• ${q}_{1}$ = heat required to cool the water from 25 °C to 0 °C
• ${q}_{2}$ = heat required to freeze the water to ice at 0 °C
• ${q}_{3}$ = heat required to cool the ice from 0 °C to -25 °C

q = q_1 + q_2 + q_3 = mc_1ΔT_1 + mΔ_text(fus)H + mc_2ΔT_2

where

${q}_{1} , {q}_{2} ,$ and ${q}_{3}$ are the heats involved in each step.
$m$ is the mass of the sample
ΔT = T_"f" -T_"i"
${c}_{1}$ = the specific heat capacity of water
${c}_{2}$ = the specific heat capacity of ice
Δ_text(fus)H = the enthalpy of fusion of ice

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

$m = \text{120 g}$
${c}_{1} = \text{4.184 J·°C"^"-1""g"^"-1}$
ΔT = "0 °C - 25 °C" = "-25 °C"

q_1 = mcΔT = 120 color(red)(cancel(color(black)("g"))) × 4.184 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-25") color(red)(cancel(color(black)("°C"))) = "-12 600 J"

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

Δ_"fus"H = "334 kJ·mol"^"-1"

q_2 = 120 color(red)(cancel(color(black)("g"))) × 334color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-40 080 kJ"

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

${c}_{2} = \text{2.09 J·°C"^"-1""g"^"-1}$
ΔT_2 = "0 °C - 25°C" = "-25 °C"

q_3 = mcΔT = 120 color(red)(cancel(color(black)("g"))) × 2.09 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-25") color(red)(cancel(color(black)("°C"))) = "-6270 J"#

$q = {q}_{1} + {q}_{2} + {q}_{3} = \text{-12 600 J" - "40 080 J" - "6270 J" = "-59 000 J}$