Newton’s Law of Cooling Calculator: Modeling How Temperature Falls Over Time
About the Author
Marko Šinko
Co-Founder & Lead Developer, AI Math Calculator
A Newton’s law of cooling calculator earns its keep in some odd places: a coroner estimating time of death from a body’s temperature, a food inspector checking whether soup passed through the 60–5 °C danger zone too slowly, a physics student on problem 4 of tonight’s homework. All three are running the same equation. An object’s temperature doesn’t fall at a steady rate — it falls fast while it’s much hotter than the room and slower and slower as the gap closes, which is why coffee goes from scalding to drinkable in ten minutes but then sits lukewarm for an hour. The calculator above solves that equation in any direction: temperature at a given time, time to reach a target temperature, or the cooling constant k when all you have is two thermometer readings. This article works each of those, with real numbers, and ends with the cases where Newton’s law quietly stops being true.
One Exponential, Three Temperatures
T(t) = Tₛ + (T₀ − Tₛ)·e^(−kt)
Three temperatures do all the work. T₀ is where the object starts, Tₛ is the ambient temperature of the surroundings, and T(t) is the reading after time t. The law itself is a claim about rates: an object loses heat in proportion to how far above ambient it sits. Written as a rate equation that’s dT/dt = −k(T − Tₛ), a linear first-order differential equation whose solution is the exponential above. Notice what actually decays: not the temperature, but the gap T − Tₛ. The gap shrinks by the same fraction in every equal slice of time. Start 68 °C above ambient, and if the gap halves in 14 minutes, it halves again in the next 14, and again in the 14 after that: 68 → 34 → 17 → 8.5.
That halving time is worth naming, because it turns k from an abstract constant into something you can feel. The half-cooling time is ln 2/k ≈ 0.693/k — the exact same relationship between decay constant and half-life that governs radioactive samples, because the math is identical: something shrinks in proportion to its own size. A k of 0.05 per minute means the gap to ambient halves every 13.9 minutes. A k of 0.005 per minute means every 139 minutes. Bigger k, faster cooling — k bundles up everything physical about the situation: surface area, air flow, insulation, what the mug is made of.
Running the Newton’s Law of Cooling Calculator on 90 °C Coffee
Take the calculator’s default scenario: coffee poured at T₀ = 90 °C into a 22 °C room, with k = 0.05 per minute. Where is it after 10 minutes?
T(10) = 22 + (90 − 22)·e^(−0.05 × 10)
T(10) = 22 + 68·e^(−0.5) = 22 + 68 × 0.6065
T(10) = 22 + 41.2 = 63.2 °C
Still too hot for most people. Drinking temperature is usually quoted around 60 °C, so flip the equation around and solve for time instead: t = (1/k)·ln((T₀ − Tₛ)/(T − Tₛ)) = 20·ln(68/38) = 20 × 0.582 = 11.6 minutes. And here’s the part the exponential forces on you: the first 20-degree drop, 90 to 70, takes 7.0 minutes — but the drop from 60 to 40, the same 20 degrees, takes 14.9. Every equal slice of temperature costs more time than the one before it. If you want the coffee drinkable sooner, the biggest lever isn’t waiting; it’s adding the milk immediately, which drops T₀ itself by 5–8 °C before the exponential clock even starts.
Finding k From Two Readings — the Step Most Problems Hide
No table anywhere lists the k for your mug on your desk. In practice k is always measured, and two thermometer readings are enough. Say the 90 °C coffee reads 70 °C five minutes later, in the same 22 °C room. Put the known values into the cooling equation and solve for the one unknown:
70 = 22 + 68·e^(−5k)
e^(−5k) = 48/68 = 0.7059
−5k = ln(0.7059) = −0.3483
k = 0.0697 per minute
The move in the middle — isolating the exponential, then taking a logarithm to pull the unknown out of the exponent — is the same one used to date samples in radioactive decay problems, just with temperatures instead of atom counts. Two warnings from grading a lot of these. First, subtract the ambient temperature from both readings before dividing; the ratio must be 48/68, never 70/90. Second, a k derived from readings five minutes apart is far more trustworthy than one derived from readings thirty seconds apart, where a ±0.5 °C thermometer error can swing k by 30% or more.
The Forensic Case: Estimating Time of Death
The classic application runs the equation backwards. A body is found at 30 °C in a 20 °C room. Living body temperature is 37 °C — that’s T₀. An investigator takes a second reading one hour later: 28 °C. Two readings, so first find k, then rewind the clock:
k = ln((30 − 20)/(28 − 20)) / 60 = ln(1.25)/60 = 0.00372 per minute
t = (1/k)·ln((37 − 20)/(30 − 20)) = ln(1.7)/0.00372
t = 0.5306/0.00372 ≈ 143 minutes
Death occurred roughly 2 hours 23 minutes before the body was found. Real forensic work layers corrections on top of this — clothing, body mass, drafts, the fact that a body isn’t a uniform lump of water — but this two-reading calculation is the published starting point, and it’s exactly what the calculator’s “Cooling constant k” mode automates. The general solving machinery behind all of these rearrangements lives in any differential equation calculator; Newton’s law is simply the special case famous enough to get its own tool.
Typical Cooling Constants — and Why Yours Will Differ
These are order-of-magnitude figures for still indoor air, useful for sanity-checking a measured value rather than replacing one. If your derived k lands wildly outside its row, re-check the readings.
| Object | Typical k (per min) | Gap halves every |
|---|---|---|
| Tea in a thin cup, breezy room | 0.08–0.12 | 6–9 min |
| Coffee in an open ceramic mug | 0.03–0.07 | 10–23 min |
| Same mug with a lid | 0.01–0.02 | 35–69 min |
| 2-liter pot of water | 0.005–0.01 | 69–139 min |
| Human body, clothed, indoors | 0.002–0.005 | 2.3–5.8 h |
The spread within each row is the whole story of k. Wind or a fan can multiply k by a factor of 5–10, because forced convection strips away the boundary layer of warm air an object builds around itself — that’s all blowing on soup does. Mass matters too: the pot of water and the teacup can sit side by side at 90 °C, but the pot holds roughly 10 times the heat behind a similar surface area, so its k is an order of magnitude smaller. Same law, same room, wildly different clocks.
Where Newton’s Law Stops Working
The law assumes heat leaves at a rate proportional to the temperature gap — an assumption that quietly fails in three common situations, always in the same direction: real objects cool faster than the prediction while hot, then fall back onto the exponential later.
- Radiation grows with T⁴, not T. Per the Stefan–Boltzmann law, a 200 °C baking tray in a 20 °C kitchen sheds radiant heat about 3.6 times as fast as a 100 °C one — a linear model predicts 2.25 times, since its temperature gap is 180° against 80°. For anything that hot the early cooling runs measurably ahead of the exponential; below about 50 °C over ambient, the linear approximation is respectable.
- Evaporation is a separate exit for heat. A steaming, uncovered drink loses up to half its heat through evaporation alone. That’s the real reason a lid seems to break the formula: it isn’t changing k slightly, it’s closing an entire second cooling channel the equation never modeled.
- Big objects don’t have one temperature. A roast out of the oven has a surface cooling fast and a core still rising — carryover cooking lifts the center 3–5 °C after removal. A single-k model only makes sense once the inside and outside have roughly equalized, typically 20–30 minutes for a large roast.
None of this makes the law useless — it makes it a model with a domain, like every model. For modest temperature gaps, covered containers, and objects small enough to have one temperature, its predictions routinely land within 1–2 °C over a half hour. The full physics, including the convective heat-transfer coefficients hiding inside k, is laid out in the Wikipedia treatment of Newton’s law of cooling, which is a good next stop if you need the version with material properties instead of a fitted constant. And the equation is symmetric, one detail worth remembering: a 4 °C soda warming in a 22 °C room follows the identical curve upside down, gap halving on the same schedule. The calculator handles that case without any change of settings — enter T₀ below Tₛ and the warming curve appears.
