SME
Principles of Biology II · Problem type
Transpiration rate from a potometer
Use this for “A shoot with 150 cm² of leaves loses 0.28, 0.30 and 0.32 mL over three 10-minute readings. What is its transpiration rate?”
Chart
0.28 mL
0.3 mL
0.32 mL
Water lost: W = 0.9 mL
Total time t = 30 min
Leaf area A = 150 cm²
Transpiration rate E = 0.2 μL/(cm²·min)
This is an example. Type your own values to start a new problem.
W₁
First reading
example · mL
W₂
Second reading
example · mL
W₃
Third reading
example · mL
W
Water lost
calculated · mL
t
Total time
example · min
A
Leaf area
example · cm²
E
Transpiration rate
calculated · μL/(cm²·min)
Formulas
W
=
W₁
+
W₂
+
W₃
0.9 = 0.28 + 0.3 + 0.32
E
=
1,000
×
W
A
×
t
0.2
=
1,000
×
0.9
150 × 30
Assumptions
•
The potometer measures water taken up; nearly all of it leaves through the leaves, so uptake stands for loss.
•
The rate is in μL per cm² of leaf per minute: 1 mL is 1000 μL.
•
Light, air flow and temperature stay steady across the readings.
Step-by-step
WE KNOW
W₁ = 0.28 mL, W₂ = 0.3 mL, W₃ = 0.32 mL, t = 30 min, A = 150 cm²
FIND
water lost (W), transpiration rate (E)
Step 1 · Find water lost (W)
Use W = W₁ + W₂ + W₃
Add the water lost over each reading.
W
=
0.28
+
0.3
+
0.32
W
=
0.9
mL
Step 2 · Find transpiration rate (E)
Use E = 1,000 × W ÷ (A × t)
Turn mL into μL, then share the water over each cm² of leaf and each minute.
E
=
1,000
×
0.9
150 × 30
E
=
1,000
×
0.9
4,500
E
=
900
4,500
E
=
0.2
μL/(cm²·min)
→
0.2
μL
of
water
leave
each
cm²
of
leaf
every
minute.
Check
Put the numbers back in. Both sides should match.
0.9 = 0.28 + 0.3 + 0.32 ✓
0.2
=
1,000
×
0.9
150 × 30
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.