Topic 6 · page 8 of 15

Graphs and cooling-curve corrections

The highest thermometer reading is too low, because the mixture is already cooling. Plot temperature against time, extend the cooling line back to the moment of mixing, and read ΔT from that.

13 min read6.8

Read first: Errors, uncertainty and assumptions

Why the highest reading is not enough 6.8

Heat is lost from the moment the reaction starts. The mixture never reaches the temperature it would have reached with perfect insulation, because it is already cooling while it warms. The highest reading on the thermometer is therefore too low, and so is the ΔH you calculate from it.

A graph fixes this. Record the temperature at regular intervals, plot the readings, and use the cooling part of the graph to work out what the temperature would have been at the instant of mixing.

Drawing the graph 6.8

MethodTemperature against time

  1. Before mixing, record the temperature of the solution every minute for about 3 minutes. This gives a steady starting line.
  2. At a known time (for example 4 min) add the second reagent. Do not take a reading at that time; you are busy adding and stirring.
  3. Start recording again at the next minute and carry on for 8 or so minutes.
  4. Plot temperature (°C) on the vertical axis and time (min) on the horizontal axis. Choose scales that make the data fill more than half of each axis, and label both axes with units.
  5. Mark each point with a small cross.
  6. Draw a straight line of best fit through the points before mixing and another through the cooling points after the maximum. Use a ruler. Do not join the dots.
  7. Extend the cooling line backwards to the time of mixing. Read the temperature where it meets that time.
  8. ΔT = the extrapolated temperature at the time of mixing − the starting temperature.

The data below are illustrative, for a reaction in a polystyrene cup. The reagent was added at 4 min.

Time / min Temperature / °C
0, 1, 2, 3 20.0 at each reading
4 reagent added, no reading
5 27.0
6 27.2
7 26.8
8 26.4
9 26.0
10 25.6
11 25.2
12 24.8
Graph of temperature in degrees Celsius against time in minutes, illustrative data. Four crosses at 20.0 degrees for 0 to 3 minutes lie on a flat line. Reagent is added at 4 minutes. Crosses at 5 to 12 minutes rise to 27.2 at 6 minutes and then fall by 0.4 degrees per minute. A best-fit line through the falling crosses is extended back to 4 minutes, where it reads 28.0, so the corrected temperature rise is 8.0 degrees. The highest reading would give only 7.2.
Illustrative data. The cooling line, extended back to 4 min, reads 28.0 °C, so the corrected ΔT is 8.0 °C.

Worked exampleReading ΔT from the cooling line

Use the data above.

  1. Starting temperature (steady line): 20.0 °C.
  2. The cooling points from 6 min to 12 min fall by 0.4 °C every minute: 27.2, 26.8, 26.4, ... 24.8.
  3. Extend the line back from 6 min to the time of mixing, 4 min. That is 2 min earlier, so the temperature would have been 27.2 + 2 × 0.4 = 28.0 °C.
  4. Corrected ΔT = 28.0 − 20.0 = 8.0 °C.
  5. The highest reading is 27.2 °C, which would give only 27.2 − 20.0 = 7.2 °C. That is 0.8 °C, or 10%, too small.

The correction matters. Suppose 0.0300 mol of reactant releases its energy in 50.0 g of solution:

ΔT / °C q = 50.0 × 4.18 × ΔT ΔH = −q ÷ 0.0300
uncorrected (highest reading) 7.2 1505 J −50.2 kJ mol⁻¹
corrected (cooling line) 8.0 1672 J −55.7 kJ mol⁻¹

Burners: a heating curve 6.8

For a combustion experiment the temperature rises while the flame is on. Record it every minute, put the flame out at a known time, and keep recording as the water cools. The cooling after the flame is out is treated in exactly the same way: extend the cooling line back to the time the flame was extinguished, and read the corrected final temperature there.

Worked exampleA corrected final temperature for a burner

Water starts at 20.0 °C. The flame is put out at 5 min. The readings at 6, 7, 8 and 9 min are 46.2, 45.8, 45.4 and 45.0 °C. (Illustrative data.)

  1. The cooling rate is (45.0 − 46.2) ÷ 3 = −0.4 °C per minute.
  2. At 5 min the line would read 46.2 + 0.4 = 46.6 °C.
  3. Corrected ΔT = 46.6 − 20.0 = 26.6 °C.

This corrects only for cooling after the flame goes out. Heat is also lost while the flame is on, and the graph cannot recover that, which is one reason a burner experiment is still far from the data-book value.

Check yourself

  1. Explain why the highest temperature reading in a mixing experiment gives a ΔT that is too small.
  2. A student takes readings before mixing, adds the reagent at 4 min, and takes no reading at 4 min. Explain why there should be no reading at 4 min.
  3. Readings for another reaction: 19.5 °C at each of 0, 1, 2 and 3 min; reagent added at 4 min; then 30.0, 29.5, 29.0 and 28.5 °C at 6, 7, 8 and 9 min. Use the cooling line to find the corrected ΔT.
  4. In question 3, 50.0 g of solution contains 0.0250 mol of the limiting reactant. Calculate the enthalpy change.
  5. State two mistakes in a graph in which the points are joined dot to dot and ΔT is taken from the top of the highest point.
  6. A student extends the cooling line back to the time of the highest reading, not to the time of mixing. State the effect on ΔH.

Answers cover these until you have tried the questions

  1. The mixture starts losing heat to the surroundings as soon as it warms. The temperature never reaches the value it would have reached with no heat loss, so ΔT, q and ΔH are all too small.
  2. The reagent is being added and stirred at 4 min, so no accurate reading can be taken. The graph uses the cooling line to find what the temperature would have been at that moment.
  3. The cooling rate is (28.5 − 30.0) ÷ 3 = −0.5 °C per minute. At 4 min, two minutes before the reading at 6 min, the temperature would be 30.0 + 2 × 0.5 = 31.0 °C. ΔT = 31.0 − 19.5 = 11.5 °C.
  4. q = 50.0 × 4.18 × 11.5 = 2403.5 J = 2.4035 kJ. 2.4035 ÷ 0.0250 = 96.14. The temperature rose, so ΔH = −96.1 kJ mol⁻¹.
  5. The points should be crosses with straight lines of best fit, not joined dot to dot. ΔT must come from the cooling line extended back to the time of mixing, not from the highest point.
  6. The extrapolated temperature would be lower than at the time of mixing, so ΔT is too small and ΔH is less exothermic than the true value.