SME
Climatology · Problem type
Warming that lags the forcing
Use this for “The equilibrium warming is 3 K and the ocean’s response time is 30 years. How much has it warmed after 30 years?”
Chart
ΔT(30) ≈ 1.896
This is an example. Type your own values to start a new problem.
ΔT (K)
=
ΔTeq (K)
(1
−
e
−
/
)
Formulas
ΔT
=
ΔTeq
×
(1
−
e
−
t
÷
τ
)
1.8964
=
3
×
(1
−
e
−30 ÷ 30
)
Assumptions
•
One ocean layer with one response time τ: the forcing steps up at t = 0 and then stays.
•
Each year closes the same fraction of the gap to equilibrium, so after one τ about 63 % of it is reached.
•
The real deep ocean adds a much slower part, so the last of the warming takes centuries.
Step-by-step
WE KNOW
ΔTeq = 3 K, τ = 30 yr, t = 30 yr
FIND
warming so far (ΔT)
Step 1 · Find warming so far (ΔT)
Use ΔT = ΔTeq × (1 − e−t ÷ τ)
e
−
t
÷
τ
is
the
share
of
the
gap
still
left,
so
1
minus
it
is
the
share
reached.
ΔT
=
3
×
(1
−
e
−30 ÷ 30
)
30
30
=
1
e
-1
=
0.3679
1 − 0.3679 = 0.6321
3 × 0.6321 = 1.896
ΔT = 1.8964 K
Check
Put the numbers back in. Both sides should match.
1.8964
=
3
×
(1
−
e
−30 ÷ 30
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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