Topic 6 · page 5 of 15
Measuring the energy transferred
A thermometer reading becomes an energy in joules through q = m c ΔT. The mass is always the mass of the water or solution that changed temperature, never the mass of the reactant.
Read first: Enthalpy changes of combustion, neutralisation and atomisation
Two experiments 6.5
To measure an enthalpy change you let the reaction heat or cool a known mass of water, and measure the temperature change. The specification limits this to two set-ups.
Substances mixed in an insulated container. For reactions in solution, such as neutralisation or the displacement of a metal. A polystyrene cup is a good insulator. A lid and a stirrer cut heat loss and even out the temperature. The thing that changes temperature is the solution.
Combustion in a calorimeter. For a fuel such as ethanol. The fuel burns in a spirit burner under a metal can of water. The thing that changes temperature is the water. You weigh the burner before and after to find the mass of fuel burned.
In both cases the energy that left the chemicals is the energy that entered the water or solution. The energy transferred, in joules, is found from the temperature change.
The expression 6.56.5(i)
In symbols, with the energy transferred written as q:
- m is the mass of the liquid whose temperature changed, in grams.
- c is its specific heat capacity, the energy needed to raise the temperature of 1 g by 1 °C. For water and dilute aqueous solutions use 4.18 J g⁻¹ °C⁻¹.
- ΔT is the final temperature minus the initial temperature, in °C. A rise is positive, a fall is negative.
Two assumptions turn volumes into masses. A dilute solution has the density of water, 1.00 g cm⁻³, so 50.0 cm³ of solution is 50.0 g. And its c is that of water. The assumptions are listed again with the sources of error.
| Mixing in a cup | Burning a fuel | |
|---|---|---|
| m | mass of the solution, in g (use 1.00 g cm⁻³) | mass of the water in the can |
| ΔT | rise (or fall) of the solution | rise of the water |
| The reactant's mass is used for | moles (next page) | moles (next page) |
Worked exampleEnergy from a temperature rise
200 g of water is heated and its temperature rises by 3.5 °C. Calculate the energy transferred.
- q = m × c × ΔT = 200 × 4.18 × 3.5
- = 2926 J, which is 2.9 kJ (2 significant figures, because ΔT has two)
Worked exampleMixing two solutions
25.0 cm³ of a solution is added to 25.0 cm³ of another in a polystyrene cup. The temperature rises from 19.5 °C to 26.3 °C. Calculate the energy transferred to the solution. (c = 4.18 J g⁻¹ °C⁻¹; density 1.00 g cm⁻³)
- The solution has total volume 25.0 + 25.0 = 50.0 cm³, so m = 50.0 g. Not 25.0 g.
- ΔT = 26.3 − 19.5 = 6.8 °C.
- q = 50.0 × 4.18 × 6.8 = 1421 J = 1.42 kJ
The size of q does not carry a sign. The sign of ΔH comes from the direction of the temperature change: a rise means the chemicals gave out heat (ΔH negative), a fall means they took it in (ΔH positive). The next page uses this.
Check yourself
- 150 g of water is heated from 18.0 °C to 26.5 °C. Calculate the energy transferred in joules.
- 50.0 cm³ of aqueous solution is cooled by a reaction and its temperature falls by 5.2 °C. Calculate the energy transferred in joules.
- A student burns a fuel under 200 g of water. The fuel burner loses 0.60 g in mass. Which of these two masses goes into q = mcΔT, and why?
- 4.20 kJ of energy is transferred to 200 g of water. Calculate the rise in temperature.
- 25.0 cm³ of acid and 25.0 cm³ of alkali are mixed. A student uses m = 25.0 g in q = mcΔT. State the effect on q and explain.
- In an experiment 100 g of solution cools from 24.0 °C to 19.5 °C. State the sign of ΔT and calculate q, giving the direction in which energy moves.
Answers cover these until you have tried the questions
- ΔT = 26.5 − 18.0 = 8.5 °C. q = 150 × 4.18 × 8.5 = 5330 J (5329.5 J).
- m = 50.0 g (1.00 g cm⁻³). q = 50.0 × 4.18 × 5.2 = 1087 J (1086.8 J), the size of the energy taken from the surroundings.
- The mass of the water (200 g). It is the water whose temperature is measured. The 0.60 g is used later to find moles of fuel, not in q.
- 4.20 kJ = 4200 J. ΔT = q ÷ (m × c) = 4200 ÷ (200 × 4.18) = 5.02 °C.
- The mass of the solution is 50.0 g, so using 25.0 g gives a q that is half the true value (so the final ΔH is also half the true value).
- ΔT = 19.5 − 24.0 = −4.5 °C. q = 100 × 4.18 × 4.5 = 1881 J. The solution lost 1881 J, which went into the reacting chemicals: the reaction is endothermic.