Newton's Law of Cooling Calculator
Inputs
| Solve for | Temperature after time |
|---|---|
| Initial temperature | 90 °C |
| Ambient temperature | 20 °C |
| Cooling constant k | 0.05 min⁻¹ |
| Target temperature | 40 °C |
| Measured temperature | 60 °C |
Visualization
Newton's Law of Cooling Calculator
Apply Newton's law of cooling T(t) = T_env + (T₀ − T_env)·e^(−kt) to find how an object's temperature approaches its surroundings. Solve for the temperature after a given time, the time to reach a target temperature, or the cooling constant k from a single measurement — with the characteristic time, the half-life and a cooling curve.
Inputs
Results
Enter a value to see results.
Details
Cooling curve
Newton's Law of Cooling
Newton's law of cooling describes how a warm object loses heat to cooler surroundings — or a cold object gains heat from warmer ones. Its central idea is simple: the speed of cooling is set by how far the object's temperature is from the temperature around it. A very hot cup of coffee cools quickly; the same cup, once it is only a little above room temperature, cools much more slowly. Written as a rate equation,
dtdT=−k(T−Tenv)the temperature difference is the engine of the process and is the cooling constant. Solving this equation gives the exponential curve this calculator uses:
T(t)=Tenv+(T0−Tenv)e−ktEnter the initial temperature , the ambient temperature and the cooling constant , and the calculator returns the temperature after a chosen time, the time needed to reach a target temperature, or itself fitted from a single measurement.
How the model works
The quantity that actually decays is not the temperature but the temperature gap . Substituting it into the rate equation gives , the same form as radioactive decay or a discharging capacitor. The gap therefore shrinks by a constant fraction in each equal slice of time:
ΔT(t)=ΔT0e−kt,ΔT0=T0−TenvAdding the ambient temperature back gives . Because only the gap decays, the object always approaches and never crosses it: coffee settles just above room temperature, and a chilled drink warms up to it but no further.
Two timescales summarise the curve. The characteristic time is the time for the gap to fall to about 37 % () of its starting value; after roughly the object is within 1 % of ambient and looks, for practical purposes, settled. The half-life is the time for the gap to halve, and it is the same at every point on the curve.
Formula
| Quantity | Symbol | Definition |
|---|---|---|
| Temperature at time | ||
| Time to reach target | ||
| Cooling constant | ||
| Characteristic time | ||
| Half-life |
The target and measured temperatures must lie between and , because the curve only ever moves from toward .
Worked example
A cup of coffee is poured at into a room at , with a cooling constant . What is its temperature after minutes?
Step 1 — Temperature gap at the start:
ΔT0=T0−Tenv=90−20=70∘CStep 2 — Decay factor:
e−kt=e−0.05×10=e−0.5≈0.6065Step 3 — Temperature after 10 minutes:
T=Tenv+ΔT0e−kt=20+70×0.6065≈62.5∘CAfter ten minutes the coffee has dropped from 90 °C to about 62.5 °C. Its characteristic time is minutes and its half-life is minutes — so every 14 minutes the remaining gap to room temperature halves.
Finding the cooling constant from a measurement
The cooling constant is a property of the object and its surroundings — surface area, material, and how freely air moves past it — and is almost always measured rather than looked up. Take one extra temperature reading at a known time and rearrange the law:
k=−t1ln(T0−TenvTm−Tenv)For the coffee above, suppose it reads after 10 minutes. Then
k=−101ln(90−2060−20)=−101ln(0.571)≈0.056 min−1Once is known, the whole future curve — and the time to reach any drinkable target — follows.
How the gap closes over time
| Elapsed time | Gap remaining | Note |
|---|---|---|
| 100% | starting gap | |
| 36.8% | one characteristic time | |
| 50% | one half-life | |
| 13.5% | ||
| 5.0% | ||
| 0.7% | practically settled |
Real-world relevance
The law is the everyday workhorse of cooling estimates. It sets how long a hot dish must rest before serving, guides the cooling schedules that keep food safe as it passes through the danger zone, and appears in forensic science, where the drop in body temperature after death gives a rough estimate of the time elapsed. Engineers use the same exponential to size the settling time of electronics and to check that a component reaches a safe temperature within a required window.
Limitations
Newton's law is an approximation that holds best when heat leaves mainly by convection into a large, steady reservoir and the temperature difference is modest. When an object is very hot, radiation takes over and heat loss follows the fourth-power Stefan–Boltzmann law rather than a straight-line one, so cooling is faster than a single constant predicts. The model also assumes the object has one uniform temperature and that the surroundings do not warm up; a thick roast with a cold core, or a small sealed box that heats up around the object, will both depart from the simple curve.
Frequently Asked Questions (FAQ)
What is Newton's law of cooling?
Newton's law of cooling states that the rate at which an object changes temperature is proportional to the difference between its temperature and that of its surroundings. In symbols, dT/dt = −k(T − T_env), whose solution is the exponential curve T(t) = T_env + (T₀ − T_env)·e^(−kt).
Because the driving force is the temperature gap, cooling is fast when the object is much hotter than its surroundings and slows as the gap closes — which is why coffee cools quickly at first and then lingers just above room temperature.
How do I find the cooling constant k?
The cooling constant is rarely known in advance; it is measured. Record the initial temperature, the ambient temperature, and one later temperature at a known time, then solve the cooling law for k: k = −ln[(T_m − T_env)/(T₀ − T_env)] / t.
For example, coffee at 90 °C in a 20 °C room that falls to 60 °C after 10 minutes gives k = −ln[(60 − 20)/(90 − 20)] / 10 min ≈ 0.056 min⁻¹. The “cooling constant” mode of this calculator does exactly this arithmetic.
When is the law a good approximation?
Newton's law works well when heat leaves mainly by convection into a large reservoir whose temperature barely changes, and when the temperature difference is modest. It assumes a single cooling constant, uniform object temperature and constant surroundings.
It is less accurate when radiation dominates — at large temperature differences heat loss follows the fourth-power Stefan–Boltzmann law rather than a linear one — or when the surroundings themselves heat up appreciably, as in a small sealed container.
Does the same formula describe warming up?
Yes. The same equation handles an object warming towards a hotter environment — a cold drink left on a table, for instance. If the initial temperature is below the ambient temperature, the term (T₀ − T_env) is negative, so the curve rises towards the ambient value instead of falling.
Either way the object approaches the surrounding temperature and never overshoots it, and the characteristic time τ = 1/k describes how quickly the gap closes in both directions.
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