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_1color(white)(mm) = "the specific heat capacity of ice" = "0.50 cal·°C"^"-1""g"^"-1"
c_3color(white)(mm) = "the specific heat capacity of water" = "1.00 cal·°C"^"-1""g"^"-1"
c_5 color(white)(mm)= "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"
bbq_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"
bbq_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"
bbq_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"
bbq_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"
bbq_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 = "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.
