The Sugar Dissolving Investigation
Four students investigate how the temperature of water affects the time taken for sugar to dissolve completely. Each takes a clean 250 mL beaker with 100 mL of water and records the time for the last crystal to disappear:
| Student | Water temperature | Sugar | Stirring | Time |
|---|---|---|---|---|
| P | 20 °C | 5 g lump | none | 210 s |
| Q | 40 °C | 5 g lump | none | 96 s |
| R | 60 °C | 10 g lump | continuous | 41 s |
| S | 80 °C | 5 g lump | none | 18 s |
They plot time against temperature, include all four points, draw a smooth curve and conclude: "The rate of dissolving doubles for every 20 °C rise in temperature." Their teacher marks the graph down, not because the numbers are wrong, but because the experiment cannot support the claim as designed.
1. Identify which data point cannot be used in the temperature comparison and list every variable the group failed to control, with the effect of each.
Read the worked answer
The 60 °C point (R) must be excluded from a temperature comparison. Two things differ from the other three runs: the mass of sugar is 10 g rather than 5 g, and the beaker was stirred continuously. A larger mass takes longer to dissolve; stirring shortens the time. The two effects are in opposite directions, so the 41 s figure cannot be compared with 210 s, 96 s or 18 s in any meaningful way.
Variables not controlled:
- Mass of sugar - more solid means more material to dissolve and a longer time.
- Stirring - agitation carries dissolved sugar away from the crystal surface and brings fresh water to it, sharply reducing the time.
- Particle size - though here all four used lump sugar, this must be explicitly stated and held constant, because powder dissolves far faster than a lump of the same mass as surface area increases.
- Volume of water - stated as 100 mL in all runs, so this one was controlled; the group should say so.
A fair test varies only temperature. The fix is straightforward: repeat the 60 °C run with 5 g of lump sugar and no stirring, and repeat each temperature two or three times and average to reduce random error.
2. The group drew a smooth curve through all four points and then stated an exact doubling rule. Explain why including R misleads the reader and what a properly drawn graph would show.
Read the worked answer
The curve is drawn through a point obtained under different conditions, so it is not a graph of "time versus temperature" at all - it mixes three variables. Because R's time is artificially short, the curve is dragged downwards in the middle of the range, producing a shape the experiment did not measure. Any statement about the 40–80 °C interval is then distorted.
A properly constructed graph would:
- plot only P, Q and S, i.e. runs differing solely in temperature;
- label both axes with quantity and unit (temperature in °C on the x-axis, time in seconds on the y-axis);
- use a scale that spreads the three points, state whether the origin is included, and draw a smooth curve rather than joining points with straight segments;
- carry error bars or note that each point is a single unrepeated measurement.
With three collinear-ish points the honest conclusion is only "as temperature rises, the time taken falls, i.e. the rate of dissolving increases". A precise doubling rule cannot be justified from three unreplicated points, one of which covers a different interval. The missing 60 °C value should be measured rather than estimated from the curve and then quoted as data.
3. State the conclusion that the valid data do support, test the doubling claim arithmetically, and suggest two improvements that would make the result more reliable.
Read the worked answer
Supported conclusion: within 20–80 °C, the rate at which sugar dissolves increases with temperature - the time needed falls from 210 s to 18 s, a factor of about 11.7. Rate is proportional to 1/time, so the rate rises by the same factor.
Testing the doubling claim: a doubling every 20 °C over three intervals would reduce the time by 2³ = 8 times, to about 26 s. The actual drop is 210 → 18 s, about 11.7 times. Looking interval by interval, 20 → 40 °C gives 210/96 ≈ 2.2, while 40 → 80 °C (two intervals, so fourfold expected) gives 96/18 ≈ 5.3. The trend is real but not an exact doubling rule; the rate of increase slows at higher temperature. The claim is therefore an over-precision, and the missing 60 °C point cannot be invented to fill the gap.
Improvements: (i) repeat each temperature at least three times with a fresh beaker of water and report the mean, discarding outlier readings only with a reason; (ii) control particle size exactly by crushing all samples and sieving to the same mesh, and use a thermometer clamped in the water (not resting on the beaker base) to read the true temperature at the moment the sugar is added.