Rate calculations for transpirationSpec 6.13
In short
A rate of transpiration or water uptake is the change in a quantity divided by the time taken, for example the distance a bubble moves in a potometer per minute. Percentage change in mass equals change in mass divided by original mass, times 100. Calculate a mean from repeats first, and always give the unit, such as mm per minute.
A rate tells you how much something changes in a unit of time. For water uptake or transpiration you can use the distance moved by a bubble in a potometer, the volume of water lost, or the change in mass of a plant.
Rate of water uptake
In a potometer, an air bubble moved 31 mm, 28 mm and 34 mm in three repeats of 10 minutes each. Calculate the mean rate of water uptake in mm per minute.
- Mean distance = (31 + 28 + 34) ÷ 3 = 93 ÷ 3 = 31 mm.
- Rate = distance ÷ time = 31 ÷ 10 = 3.1 mm per minute.
Answer: 3.1 mm per minute
Percentage loss of mass
A leafy shoot had a mass of 25.0 g at the start of an investigation and 23.0 g at the end. Calculate the percentage loss of mass.
- Loss of mass = 25.0 − 23.0 = 2.0 g.
- Percentage loss = 2.0 ÷ 25.0 × 100 = 8%.
Answer: 8%
Always include the unit with a rate, for example mm per minute or cm³ per hour. Calculate a mean first if you are given repeats.
Quick check
What does a potometer measure?
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The rate of water uptake of a leafy shoot.
Why does moving air increase the rate of transpiration?
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It carries away water vapour, keeping a steep concentration gradient so water vapour diffuses out faster.
A bubble in a potometer moves 24 mm in 8 minutes. What is the rate?
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24 ÷ 8 = 3 mm per minute.
Why does a higher temperature increase the rate of water uptake?
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Water evaporates from the leaf cells faster and water vapour diffuses out faster, so more water is pulled up the xylem.
Written and checked against the Edexcel GCSE Combined Science (1SC0) specification · Updated October 2026