Heating and Cooling Curves
Heating and Cooling Curves
In the tutorial on thermochemistry, the relation between the amount of heat absorbed or released by a substance, q, and its accompanying temperature change, ΔT, was introduced:
\(q=mc\text{Δ}T\)
where m is the mass of the substance and c is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. The figure below shows a typical heating curve.
Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water’s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner.
This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.
A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions.
Example
Total Heat Needed to Change Temperature and Phase for a Substance
How much heat is required to convert 135 g of ice at −15 °C into water vapor at 120 °C?
Solution
The transition described involves the following steps:- Heat ice from −15 °C to 0 °C
- Melt ice
- Heat water from 0 °C to 100 °C
- Boil water
- Heat steam from 100 °C to 120 °C
The heat needed to change the temperature of a given substance (with no change in phase) is: q = m\(×\)c\(×\) ΔT (see previous tutorial on thermochemistry). The heat needed to induce a given change in phase is given by q = n\(×\) ΔH.
Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:
\({q}_{\text{total}}={\left(m\text{⋅}c\text{⋅}\text{Δ}T\right)}_{\text{ice}}+n\text{⋅}\text{Δ}{H}_{\text{fus}}+{\left(m\text{⋅}c\text{⋅}\text{Δ}T\right)}_{\text{water}}+n\text{⋅}\text{Δ}{H}_{\text{vap}}+{\left(m\text{⋅}c\text{⋅}\text{Δ}T\right)}_{\text{steam}}\)
\(\begin{array}{}\\ \\ =\left(\text{135 g}\text{⋅}\text{2.09 J/g}\text{⋅}\text{°}\text{C}\text{⋅}15\text{°}\text{C}\right)+\left(135\text{⋅}\frac{\text{1 mol}}{18.02\phantom{\rule{0.2em}{0ex}}\text{g}}\text{⋅}\text{6.01 kJ/mol}\right)\phantom{\rule{0.2em}{0ex}}\\ +\left(\text{135 g}\text{⋅}\text{4.18 J/g}\text{⋅}\text{°}\text{C}\text{⋅}100\text{°}\text{C}\right)+\left(\text{135 g}\text{⋅}\frac{\text{1 mol}}{18.02\phantom{\rule{0.2em}{0ex}}\text{g}}\text{⋅}\text{40.67 kJ/mol}\right)\\ +\left(\text{135 g}\text{⋅}\text{1.84 J/g}\text{⋅}\text{°}\text{C}\text{⋅}20\text{°}\text{C}\right)\\ =\text{4230 J}+\text{45.0 kJ}+\text{56,500 J}+\text{305 kJ}+\text{4970 J}\end{array}\)
Converting the quantities in J to kJ permits them to be summed, yielding the total heat required:
\(=4.23\phantom{\rule{0.2em}{0ex}}\text{kJ}+\text{45.0 kJ}+\text{56.5 kJ}+\text{305 kJ}+\text{4.97 kJ}=\text{416 kJ}\)
This lesson is part of:
Liquids and Solids