Cooling Time Calculator (Lumped Capacitance)
Temperature of a part after a given time in air or a bath, plus its thermal time constant and Biot number.
Results
What this tool does
A hot part left in cooler surroundings loses temperature on an exponential curve, not a straight line: it sheds the first half of the gap quickly and the last part of it very slowly. The time constant tau tells you the whole story — after one tau the part has covered 63% of the way to ambient, after three tau it is 95% there. The Biot number is the sanity check on whether treating the part as a single uniform lump is honest.
Formula
T(t) = T∞ + (T₀ − T∞) e^(−t/τ) , τ = ρ c Lc ÷ h
Variables
| Symbol | Meaning | Unit |
|---|---|---|
hc | Overall heat transfer coefficient | W/(m²·K) |
lc | Characteristic length (V/A) | mm |
ro | Density | kg/m³ |
cp | Specific heat capacity | J/(kg·K) |
kk | Thermal conductivity | W/(m·K) |
t0 | Initial temperature | °C |
ti | Surrounding temperature | °C |
ts | Time | s |
TT | Final temperature | °C |
TA | Time constant | s |
BI | Biot number | — |
TH | Half-life | s |
Worked example
- Overall heat transfer coefficient25 W/(m²·K)
- Characteristic length (V/A)5 mm
- Density7850 kg/m³
- Specific heat capacity470 J/(kg·K)
- Thermal conductivity50 W/(m·K)
- Initial temperature200 °C
- Surrounding temperature25 °C
- Time300 s
- Final temperature141.54 °C
- Time constant737.9 s
- Biot number0.00250
- Half-life511.5 s
Limitations
- Mixing units is the most common source of error. Convert every input to the units shown next to each field before calculating.
- The formula assumes ideal conditions: no friction losses, no air resistance and no efficiency losses unless you enter them.
- The result is an estimate based only on the values you type. Real situations often include factors this calculator does not know about.
Frequently asked questions
What does the Biot number have to be for this to be valid?
Below 0.1. The whole method assumes the body is at one uniform temperature throughout, which only holds when conduction inside it is far easier than convection off its surface. A small steel part in still air satisfies this comfortably; a thick ceramic block quenched in water does not, and then you need a full transient conduction solution instead.