Heating Curves


A heating curve of a substance shows the relationship of temperature, state of matter, and heat (when added over time). Substances undergo phase transitions at their melting and boiling points.

Consider a substance in the solid state below its freezing point. To convert the substance to a gas above the boiling point, the following must occur:

  • Heat solid to its melting point
  • Melt the solid from solid to liquid (fusion)
  • Heat liquid to its boiling point
  • Vaporize liquid to a gas (vaporization)
  • Heat gas to the final temperature

Below is an illustration of the heating process for a solid at some initial temperature (Tinitial) to a gas at some final temperature (Tfinal).

A general heating curve

Substances experience a temperature change upon heating except for when a phase transition occurs. During a phase transition, the temperature does not change (ΔT = 0).

When a certain amount of substance experiences a temperature change, the amount of heat (q) required to change the temperature of that substance is given as:

\[\begin{equation} q = mc\Delta T \end{equation}\]

where q is the amount of heat, m is the mass of the substance, c is the specific heat of the substance for a given phase, and ΔT is the temperature change.

The amount of heat required for a certain amount of substance undergoing a phase transition is

\[\begin{equation} q = n\Delta H \end{equation}\]

where q is the amount of heat, n is the number of moles of the substance, and ΔH is the enthalpy of that phase transition (generally given in kJ mol–1).

Generally, the specific heats of a substance in some phase, the enthalpy of fusion, and the enthalpy of vaporization are given for these types of problems that require the determination of q.


Substance data

Enthalpies of fusion and vaporization (in kJ mol–1) for some substances measured at their respective (approximate) normal freezing and boiling points (Tf and Tb in oC). Specific heats (c) are given in J g–1 oC–1.

Substance Tf Tb ΔHfus ΔHvap cs cl cg
water 0 100 6.02 44.01 2.09 4.184 1.865
benzene 5.53 80.1 9.87 30.77 1.1 1.741 1.16
methanol −97.6 64.7 3.18 38.2 ··· 2.531 1.376
acetone −95 56 5.7 31.3 1.653 2.409 1.283
butane −138 −1 4.66 21 ··· 2.062 1.639
ammonia −77.73 −33.34 5.66 23.35 ··· 2.061 1.703
propane 188 −42 1.96 15.7 ··· 2.025 1.669


Practice Problems


Problem 1

How much heat (in kJ mol–1) is required to heat 300 g of water from –40 oC to 130 oC?

Solution

\[\begin{align*} q_1 &= mc_{solid}\Delta T \\ &= 300~\text{g} \left ( 2.09~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 0~^{\circ}\text{C} - -40~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 25.08~\mathrm{kJ} \\[2ex] q_2 &= n\Delta H_{\mathrm{fus}} \\ &= 300~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) \left ( \dfrac{6.02~\mathrm{kJ}}{\mathrm{mol}} \right ) \\ &= 100.2~\mathrm{kJ} \\[2ex] q_3 &= mc_{liquid}\Delta T \\ &= 300~\text{g} \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 100~^{\circ}\text{C} - 0~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 125.5 ~\mathrm{kJ} \\[2ex] q_4 &= n\Delta H_{\mathrm{vap}} \\ &= 300~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) \left ( \dfrac{44.01~\mathrm{kJ}}{\mathrm{mol}} \right ) \\ &= 732.7~\mathrm{kJ} \\[2ex] q_5 &= mc_{gas}\Delta T \\ &= 300~\text{g} \left ( 1.84~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 130~^{\circ}\text{C} - 100~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\ &= 16.56 ~\mathrm{kJ} \\[2ex] q_{\mathrm{tot}} &= 25.08 + 100.2 + 125.5 + 732.7 + 16.56 = 1000~\mathrm{kJ} \end{align*}\]


Problem 2

How much heat (in kJ mol–1) is required to heat 2.45 mol of water from –40 oC to 130 oC?

Solution

\[\begin{align*} q_1 &= mc_{solid}\Delta T \\ &= 2.45~\mathrm{mol} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \left ( 2.09~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 0~^{\circ}\text{C} - -40~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 3.691~\mathrm{kJ} \\[2ex] q_2 &= n\Delta H_{\mathrm{fus}} \\ &= 2.45~\mathrm{mol}\left ( \dfrac{6.02~\mathrm{kJ}}{\mathrm{mol}} \right ) \\ &= 14.75~\mathrm{kJ} \\[2ex] q_3 &= mc_{liquid}\Delta T \\ &= 2.45~\mathrm{mol} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 100~^{\circ}\text{C} - 0~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 18.47 ~\mathrm{kJ} \\[2ex] q_4 &= n\Delta H_{\mathrm{vap}} \\ &= 2.45~\mathrm{mol}\left ( \dfrac{44.01~\mathrm{kJ}}{\mathrm{mol}} \right ) \\ &= 107.8~\mathrm{kJ} \\[2ex] q_5 &= mc_{gas}\Delta T \\ &= 2.45~\mathrm{mol} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \left ( 1.84~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 130~^{\circ}\text{C} - 100~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 2.45 ~\mathrm{kJ} \\[2ex] q_{\mathrm{tot}} &= 3.691 + 14.75 + 18.47 + 107.8 + 2.45 = 147.2~\mathrm{kJ} \end{align*}\]


Problem 3

How much heat (in kJ mol–1) is required to heat 300.0 g of water from –40 oC to 70 oC?

Solution

\[\begin{align*} q_1 &= mc_{solid}\Delta T \\ &= 300~\text{g} \left ( 2.09~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 0~^{\circ}\text{C} - -40~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 25.08~\mathrm{kJ} \\[2ex] q_2 &= n\Delta H_{\mathrm{fus}} \\ &= 300~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) \left ( \dfrac{6.02~\mathrm{kJ}}{\mathrm{mol}} \right ) \\ &= 100.2~\mathrm{kJ} \\[2ex] q_3 &= mc_{liquid}\Delta T \\ &= 300~\text{g} \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 70~^{\circ}\text{C} - 0~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 87.86 ~\mathrm{kJ} \\[2ex] q_{\mathrm{tot}} &= 25.08 + 100.2 + 87.86 = 213.14~\mathrm{kJ} \end{align*}\]


Problem 4

How much heat (in kJ mol–1) is required to heat 2.45 mol of water from –40 oC to 70 oC?

Solution

\[\begin{align*} q_1 &= mc_{solid}\Delta T \\ &= 2.45~\mathrm{mol} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \left ( 2.09~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 0~^{\circ}\text{C} - -40~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 3.961~\mathrm{kJ} \\[2ex] q_2 &= n\Delta H_{\mathrm{fus}} \\ &= 2.45~\mathrm{mol}\left ( \dfrac{6.02~\mathrm{kJ}}{\mathrm{mol}} \right )\\ &= 14.75~\mathrm{kJ} \\[2ex] q_3 &= mc_{liquid}\Delta T \\ &= 2.45~\mathrm{mol} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 70~^{\circ}\text{C} - 0~^{\circ}\text{C} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\ &= 12.93 ~\mathrm{kJ} \\[2ex] q_{\mathrm{tot}} &= 3.961 + 14.75 + 12.93 = 31.6~\mathrm{kJ} \end{align*}\]


Handouts


Acetone


Timed Assessments


Water

VIDEO

Answers
  1. 122 kJ
  2. 339 kJ
  3. 172 kJ

Benzene

VIDEO

Answers
  1. 28.1 kJ
  2. 90.8 kJ
  3. 117 kJ
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