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Climatology
Climate models
Use this for “CO₂ doubles from 280 ppm to 560 ppm. With a sensitivity of 0.8 K per W/m², what are the forcing and the equilibrium warming?”
Chart
ΔF(560) ≈ 3.708
This is an example. Type your own values to start a new problem.
ΔF (W/m²)
=
5.35
ln
C (ppm)
C₀ (ppm)
λ
Climate sensitivity
example · K/(W/m²)
ΔT
Equilibrium warming
calculated · K
Formulas
ΔF
=
5.35
×
ln(
C
C₀
)
3.7083
=
5.35
×
ln(
560
280
)
ΔT
=
λ
×
ΔF
2.9667 = 0.8 × 3.7083
Assumptions
•
Forcing grows with the log of CO₂, ΔF = 5.35 ln(C ÷ C₀), so each doubling adds the same 3.7 W/m².
•
The sensitivity λ includes the feedbacks (water vapour, ice, clouds); 0.8 K per W/m² gives 3 K a doubling.
•
ΔT is the equilibrium warming, reached only once the deep ocean has warmed too.
Step-by-step
WE KNOW
C₀ = 280 ppm, C = 560 ppm, λ = 0.8 K/(W/m²)
FIND
forcing (ΔF), equilibrium warming (ΔT)
Step 1 · Find forcing (ΔF)
Use ΔF = 5.35 × ln(C ÷ C₀)
The forcing follows the natural log of how many times the CO₂ has grown.
ΔF
=
5.35
×
ln(
560
280
)
560
280
=
2
ln(2) = 0.6931
ΔF
=
3.70834
W/
m
2
Step 2 · Find equilibrium warming (ΔT)
Use ΔT = λ × ΔF
Each
W/
m
2
of
forcing
warms
the
surface
by
λ
kelvins
once
it
settles.
ΔT = 0.8 × 3.70834
ΔT = 2.9667 K
Check
Put the numbers back in. Both sides should match.
3.70834
=
5.35
×
ln(
560
280
)
✓
2.9667 = 0.8 × 3.70834 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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