Explain changing temperature with an energy balance

Follow the first 200 joules

Choose standard hardware at gain 6 W/K. Initially the true and sensed temperatures are 20°C. The 30°C target makes the controller request 10 + 6 × 10 = 70 W, but the heater limit permits only 20 W. That applied input lasts ten seconds, supplying 200 J. Predict whether the device stores all 200 J before looking at its first receipt.

The device has heat capacity C = 100 J/K. If no energy escaped, 200 J would raise it by 2 K. The actual model first endpoint is about 21.9032516393°C. Once the body becomes warmer than its 20°C surroundings, heat transfer outward begins. The smaller rise is therefore consistent with energy conservation, not evidence that energy disappeared.

Read the balance in words

The model says C dT/dt = P − G(T − Ta). T is the uniform body temperature, Ta the maintained ambient temperature, P the applied heater power and G the total conductance to ambient. C is in J/K, G in W/K and time in seconds. The left side is the rate of stored-energy change; the right side is input power minus outward transfer rate. You do not need calculus to use the interval ledger.

For the first interval, stored-energy change is C times the endpoint rise: 100 × 1.9032516393 = 190.32516393 J. Ambient transfer is input minus storage, approximately 9.67483607 J. The account closes: 200 = 190.325 + 9.675, within displayed rounding. Use full precision when checking the underlying calculation; do not expect rounded columns to close perfectly.

At later intervals the controller selects a new input from the updated sensed temperature. Do not multiply the first 20 W by 120 seconds to infer the entire run’s energy. Sum the twelve applied-power-times-ten-second receipts. Requested power above the actuator limit is a command, not additional delivered energy.

Understand the exact interval solution

During one interval, P and Ta are fixed. The corresponding equilibrium temperature is Teq = Ta + P/G. Define τ = C/G, the thermal time constant in seconds. The exact update is Tnext = Teq + (Tcurrent − Teq) exp(−Δt/τ). The exponential factor tells what fraction of the gap to that interval’s equilibrium remains.

For the first standard interval, Teq = 20 + 20/1 = 40°C, τ = 100/1 = 100 s, and Δt = 10 s. Hence Tnext = 40 − 20 exp(−0.1) ≈ 21.903°C. The controller changes P after ten seconds, so 40°C is not necessarily the eventual controlled temperature. A time constant is also not a deadline at which equilibrium is exactly reached.

Capture insulated hardware at the same gain and inspect its first receipt. It has the same capacity, initial state and applied 20 W, but a smaller conductance. Explain why less energy escapes during the interval. Do not attribute the difference to a larger heater or a changed start.

A separate passive-cooling experiment on paper

Now consider a different, explicitly hypothetical experiment: the body starts at 40°C, ambient remains 20°C, and the heater is off throughout. This is not a selectable mode of the widget. The exact expression reduces to T(t) = 20 + 20 exp(−t/τ).

After 100 seconds, standard hardware with τ = 100 s would reach about 27.358°C. Insulated hardware with τ = 200 s would reach about 32.131°C. Insulation slows the loss of the initial thermal excess. It has not produced extra energy. COMSOL’s lumped-capacitance explanation provides the general modeling context; these numerical examples are original calculations, not cooling observations.

Transfer: change the heat capacity

In the passive paper example, keep G = 1 W/K and double C from 100 to 200 J/K. Predict the time constant and temperature after 100 seconds. Does matching one temperature prove the hardware is identical to the insulated design?

Check the capacity transfer

The time constant becomes 200 s and temperature is about 32.131°C. The cooling curve matches the insulated example because C/G is equal, but capacity and conductance differ. Stored energy for the same temperature rise is twice as large when C doubles. One temperature curve cannot identify both parameters independently without additional information.

Return to temperature and heat for capacity and stored energy, or continue with sensitivity and calculation error. Keep a model result distinct from physical evidence of spatially uniform temperature.