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Lumped Capacitance Cooling Curve

T=T∞+(T0−T∞)e−t/τT = T_\infty + (T_0 - T_\infty) e^{-t/\tau}

Worked example: 200 C block, tau 405 s, 25 C air, after 300 s → 108.4 C — press Try an example to run it live, then adjust anything.

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Lumped Capacitance Cooling Curve explained

TT∞T0τt

When Bi < 0.1 the interior of a body stays essentially uniform, an energy balance gives ρVc dT/dt = −hA(T − T∞), and the solution is a pure exponential approach to ambient. Every point on the curve is the same fraction of the remaining gap, which is why the answer never depends on how you got there. Worked example: an aluminium block at 200 °C with τ = 405 s, cooling in 25 °C air, after 300 s sits at 25+175×e−0.741=108.425 + 175 \times e^{-0.741} = 108.4 °C.

Run it the other way and it becomes the field measurement everyone actually uses: log a cooling curve, read the time to fall from 300 °F to 150 °F in 70 °F air, and τ = 634 ÷ ln(230/80) drops out — no need to know ρ, V, c, h or A at all. Two traps. First, temperature differences are what matter, so the ratio inside the logarithm works in any consistent scale, but the temperatures you type must be absolute or Celsius, not differences. Second, the model dies quietly when h is not constant: a body that starts by boiling its quench fluid and finishes in ordinary convection has two different time constants, and forcing one exponential through that data gives an h that describes neither regime.

Lumped Capacitance Cooling Curve formula

T=T∞+(T0−T∞)e−t/τT = T_\infty + (T_0 - T_\infty) e^{-t/\tau}
Where
  • TT= Temperature at time t (°C)
  • T0T_0= Initial temperature (°C)
  • T∞T_\infty= Fluid temperature (°C)
  • tt= Elapsed time (s)
  • τ\tau= Thermal time constant (s)

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