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SCIENCE · INVESTIGATE AND EXPLAIN

Separate response time from equilibrium

Compare adjustment speeds and conditional targets before and after a finite forcing interval.

Learning outcome: Use capacity and net damping to calculate response time and explain incomplete approach to equilibrium.

Starting recap

The sustained-forcing target is F/λ; response time is C/λ. The actual forcing stops at year 25.

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Try

At feedback 1.2, depths 50 and 200 m share a 2 K target but have response times about 3.248 and 12.992 years.

Open the experiment

Predict

Can two reservoirs have the same target but different temperatures when forcing is removed?

Experiment

Compare 50 and 200 m at feedback 1.2. Record response time, initial rate, conditional target and year-25 anomaly.

Store energy and trace a feedback response

Worked teaching model · synthetic values · Synthetic uniform linear reservoir: C = 4.1 million × depth J/m²/K, baseline damping 3.2 W/m²/K. An imposed +4 W/m² forcing acts until year 25, then is removed. Temperature starts at zero anomaly. Positive feedback f adds fθ to net input; total damping 3.2 − f stays positive. Exact piecewise solutions, not time-stepped estimates. Effective depth and coefficients are teaching assumptions, not fitted climate parameters or measured ocean properties.

Reset keeps your prediction, captures and practice history.

Change one parameter and compare the solid selected response with the dashed no-added-feedback reference (f = 0) at the same reservoir depth. The full forcing pulse and recovery are shown without animation. A year is 31,557,600 seconds in this model. Values displayed as zero may be rounded small values.

Storage, damping and the target if forcing were maintained
QuantitySelected responseNo-added-feedback reference
Areal heat capacity (MJ/m²/K)205.000205.000
Total damping (W/m²/K)2.0003.200
Response time (years)3.2482.030
Initial warming rate (K/year)0.6160.616
Target if +4 W/m² persisted (K)2.0001.250
Actual peak at removal, year 25 (K)1.9991.250
Temperature anomaly (K) · solid selected / dashed f = 0 reference+4 W/m² for years 0–25; zero thereafter · removal at the dotted line00.511.522.533.50255075100Elapsed model years · synthetic trajectories, not historical observations
Net energy input (W/m²) · solid selected / dashed f = 0 reference+4 W/m² for years 0–25; zero thereafter · removal at the dotted line-4-20240255075100Elapsed model years · synthetic trajectories, not historical observations
Stored energy anomaly (MJ/m²) · solid selected / dashed f = 0 reference+4 W/m² for years 0–25; zero thereafter · removal at the dotted line05001000150020000255075100Elapsed model years · synthetic trajectories, not historical observations

At the vertical removal line, temperature and stored energy are continuous. Net input drops by exactly 4 W/m²; the vertical segment is a prescribed jump, not a numerical overshoot. A positive temperature anomaly can coexist with negative net input while stored energy is being lost. The finite pulse does not have to reach the sustained-forcing target before it ends.

Signed fluxes and storage at selected times, including both sides of removal
Model yearθ (K)Forcing F (W/m²)Baseline −3.2θ (W/m²)Feedback fθ (W/m²)Net N (W/m²)Storage E (MJ/m²)
00.0004.0000.0000.0004.0000.000
51.5714.000-5.0271.8850.858322.051
101.9084.000-6.1062.2900.184391.134
25 just before1.9994.000-6.3972.3990.002409.814
25 just after1.9990.000-6.3972.399-3.998409.814
500.0010.000-0.0030.001-0.0020.186
1000.0000.0000.0000.0000.0000.000

The feedback loop is explicit: temperature anomaly θ changes the flux fθ; that flux changes net energy input N; N changes stored energy and therefore θ. Here positive f reinforces the imposed warming and negative f opposes it. The baseline response −3.2θ is always restoring. Total damping λ = 3.2 − f remains positive across every control setting, so positive added feedback does not mean runaway growth.

Energy accounting is C dθ/dt = N = F − 3.2θ + fθ and E = Cθ. C is heat capacity per area, θ is temperature change in kelvin, F and N are power per area, and t is time in seconds. The derivative means rate of temperature change. For constant forcing, target temperature is F/λ and response time is τ = C/λ. The warming solution is θ = (4/λ)[1 − exp(−t/τ)]; after removal it is θ = θ₂₅ exp[−(t − t₂₅)/τ], with all times in consistent units. The exponential gives the fraction of the remaining adjustment. The graphs sample these exact solutions; graph spacing is not an integration step.

The reference has no added feedback, but retains baseline radiative damping. Depth is an effective water-equivalent storage parameter, not actual ocean depth or a measured mixed layer. The prescribed linear response resolves no cloud, ice, water-vapor or circulation process. There are no multiple reservoirs, geography, seasonal variation, noise or gas-concentration mapping. Removing this imposed forcing is not equivalent to stopping CO₂ emissions. The 4 W/m² pulse and coefficient choices are teaching assumptions, not an estimate of real-world equilibrium climate sensitivity.

NASA's energy-budget explanation provides context for storage and delayed temperature response. IPCC AR6 Chapter 7 discusses forcing and radiative feedbacks; its net-feedback sign convention differs from the positive-damping λ used here.

Response time 3.248 years; sustained-forcing target 2.000 K. At year 25 the anomaly is 1.999 K.

Stored energy at removal 409.814 MJ/m². Net flux jumps from 0.002 to -3.998 W/m² without a temperature jump. The forcing is removed, not maintained forever; the long-time post-pulse target is zero anomaly. This is not a CO₂-emissions or real climate-sensitivity forecast.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Separate the destination from the pace

Compare effective depths 50 and 200 m while holding the feedback coefficient at 1.2 W m⁻² K⁻¹. Both receive the same imposed forcing and have the same radiative response at a given temperature anomaly. Predict whether they must nevertheless follow different temperature trajectories.

The deeper effective reservoir stores more energy per degree, so it adjusts more slowly. If the 4 W/m² forcing were maintained indefinitely, both cases would approach the same equilibrium anomaly. The actual experiment removes forcing after 25 years, so that sustained-forcing equilibrium is a conditional target, not a guaranteed endpoint.

A simple energy balance links storage with radiative response. The distinction between forcing, response and energy accumulation also appears in IPCC AR6 Chapter 7. The coefficients and reservoir here are teaching choices, not estimated climate-system parameters.

Find the conditional equilibrium

Write λ = 3.2 − f for net damping, in W m⁻² K⁻¹. Across the controls λ remains positive. During the imposed forcing, net flux is N = 4 − λθ. At equilibrium N = 0, so the conditional target is θeq = 4/λ.

At f = 1.2, λ = 2 and θeq = 2 K. Heat capacity does not appear in this equilibrium equation. It controls how quickly energy changes temperature, rather than the temperature at which the specified incoming and outgoing changes balance.

On paper, compare H = 50 m and H = 200 m. Write the same target beside both, then their different capacities: 205 and 820 million J m⁻² K⁻¹. This separates a balance condition from a storage property before the curve adds visual detail.

Interpret the response time

The response time is τ = C/λ, initially in seconds. Divide by 31,557,600 to express it in model years. At the default depth and feedback, τ is about 3.248 years. At 200 m with the same feedback, τ is four times larger, about 12.992 years.

While forcing remains active, θ(t) = θeq[1 − exp(−t/τ)]. The exponential starts at one, making θ initially zero, then decreases toward zero. After one response time, the temperature has completed about 63.2% of the approach to its target; it has not completed the adjustment. After three response times, about 95% is complete.

You can reason with these fractions without deriving the exponential. Draw a target line and mark 63.2% and 95% of its height. Predict where each depth’s trajectory lies at the same elapsed time. The interface keeps time fixed across cases so the comparison concerns response rather than mismatched observation dates.

Inspect an incomplete adjustment

At year 25, the 50 m, f = 1.2 case is about 1.9991 K above its initial state. The 200 m case is about 1.7080 K. Both have a 2 K sustained-forcing target, but the larger reservoir has not approached it as closely before removal.

After removal, the target becomes zero and each trajectory decays toward it. The larger reservoir can remain warmer later even though it warmed less during the forcing interval. This is a consequence of storage and response time, not a reversal of energy conservation.

Capture both H/f recipes and record initial warming rate, τ, conditional target and year-25 temperature. Explain why no single one of these quantities completely describes the response. A rapid initial rate is not itself a large equilibrium sensitivity.

The model has one reservoir and constant coefficients. It cannot represent multiple ocean adjustment times, geographic variation or exchange between a surface layer and deep ocean. Its effective depth is not an inferred real mixing depth.

Transfer by changing capacity

A stable model has a response time of four years and a sustained-forcing target of 3 K. Its heat capacity is tripled with forcing and damping unchanged. Predict the new response time, target and initial warming-rate ratio.

Check the response-time transfer

The response time becomes twelve years, the target remains 3 K, and the initial warming rate becomes one third as large. Capacity changes the pace of adjustment, not the equilibrium balance under these fixed assumptions.

Apply

Tripling capacity at unchanged forcing and damping changes the response time and target to:

Public practice checks a specific claim. It does not establish independent mastery.

Explain why the target is conditional and why changing capacity alters adjustment time without changing that balance.

Continue

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