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
#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"#
