There are three separate heats involved in this problem:
- q_1q1 = heat required to cool the water from 25 °C to 0 °C
- q_2q2 = heat required to freeze the water to ice at 0 °C
- q_3q3 = 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
bbq_1
m = "120 g"
c_1 = "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"
bbq_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"
bbq_3
c_2 = "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 = "-12 600 J" - "40 080 J" - "6270 J" = "-59 000 J"
