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Body and room
A body at one uniform temperature loses heat to the room. Newton's law gives an exponential approach with time constant C/UA; radiation, a convection exponent and a heater change the curve and the final temperature.
Model checks
The science
What is modelled
A body so small or so conductive that its temperature is uniform (a lumped capacity C = m c) exchanges heat with a room at temperature Ta. Heat leaves by convection and conduction with conductance UA, optionally with a power-law dependence on the temperature excess (turbulent convection), and by thermal radiation. A heater can supply power P. The temperature follows a single first-order ordinary differential equation.
Equations
C dT/dt = P - UA (|ΔT|/ΔT0)n (T - Ta) - ε σ A (T4 - Ta4)with temperatures in kelvin in the radiation term.- Plain Newton case (n = 0, ε = 0, P = 0):
T(t) = Ta + (T0 - Ta) e-t/τ,τ = C/UA. - With a heater: the equilibrium is Ta + P/UA (without radiation); in general it is the root of the heat balance.
Checks run in the page
- RK4 against the exact exponential in the Newton case, after three time constants.
- The equilibrium by bisection against a long simulation.
- The energy balance C(T - T0) = integral of (heater - convection - radiation).
Honest simplifications
A uniform body temperature (the Biot number must be small), a constant room temperature and constant material properties, a grey radiating surface that sees only the room, and no evaporation or phase change. A mug of cocoa also loses heat by evaporation, which this model leaves out.
What to try
- Press Mug with radiation: the temperature curve drops under the grey Newton curve.
- Press Water bath with a heater and note the equilibrium above the room temperature.
- Raise the convection exponent: the cool-down becomes slower near the end (the log graph bends).