A typical heating curve for a compound like #"SO"_2# is shown below.

(From SlideShare)

The melting point of #"SO"_2# is -73.15 °C (200.00 K), and the boiling point is -10.0 °C (263.1 K).

In this problem, we are starting with the liquid at point **B** , warming it to its boiling point **C**, and then evaporating it to point **D**.

Thus, there are two separate heat transfers involved:

- #q_1# = heat required to warm the liquid from from 200.5 K to 263.1 K (Point
**B** to **C** in the diagram)
- #q_2# = heat required to evaporate the liquid at 263.1 K (Point
**C** to **D**)

#q = q_1 + q_2 = mCΔT + nΔ_text(vap)H#

where

#q_1# and #q_2# are the heats involved in each step

#mcolor(white)(ml) = "the mass of the sample"#

#Ccolor(white)(ml) = "the specific heat capacity of SO"_2 = "1.36 J·°K"^"-1""g"^"-1"#

#ΔTcolor(white)(l) = T_"f" -T_"i"#

#Δ_text(fus)H = "the molar enthalpy of evaporation of SO"_2 = "24.9 kJ·mol"^"-1"#

#bbq_1#

#ΔT = "263.1 K - 200.5 K" = "62.6 K"#

#q_1 = mCΔT = 95.1 color(red)(cancel(color(black)("g"))) × 1.36 color(white)(l)"J"·color(red)(cancel(color(black)( "K"^"-1""g"^"-1"))) × 62.6 color(red)(cancel(color(black)("K"))) = "8096 J" = "8.096 kJ"#

#bbq_2#

#n = 95.1 color(red)(cancel(color(black)("g SO"_2))) × "1 mol SO"_2/(64.06 color(red)(cancel(color(black)("g SO"_2)))) = "1.485 mol SO"_2#

#q_2 = 1.485 color(red)(cancel(color(black)("mol"))) × "24.9 kJ"·color(red)(cancel(color(black)("mol"^"-1"))) = "36.97 J"#

#q = q_1 + q_2 = "(8.096 + 36.97) kJ" = "45.1 kJ"#

The process requires 45.1 kJ of heat energy.