Newton’s Law of Cooling Calculator (temperature over time)

Understanding the Newton's law of cooling calculator — concept and principle

This calculator predicts how the temperature of an object changes over time as it cools toward (or warms toward) its surroundings: a cup of coffee, a cooling casting, food left on a counter, or a body in a forensic estimate. It gives the temperature after a chosen time and the time needed to reach a target temperature.

The gap to room temperature shrinks by the same fraction every 10 minutes; 90 °C coffee is 62.46 °C after 10 min.

How it is calculated

T(t) = T_a + (T₀ − T_a) · e^(−kt)

T₀: starting temperature (°C); T_a: ambient temperature (°C).
k: cooling constant in 1/min. It equals hA/(mc): the convective coefficient times the surface area, divided by mass times specific heat.
t: elapsed time (min).

Solving for time gives t = ln[(T₀ − T_a)/(T_target − T_a)] / k.

The principle

Newton observed that the rate of cooling is proportional to the temperature difference with the surroundings: dT/dt = −k(T − T_a). A large difference drives fast cooling; as the object approaches ambient temperature the driving force shrinks and cooling slows. The solution of that differential equation is an exponential decay of the difference. The quantity 1/k is the time constant: after each 1/k the remaining difference falls to about 37% (1/e). The model assumes the object has a uniform internal temperature (small Biot number) and that k stays constant.

Worked example

Inputs: T₀ = 90 °C, T_a = 20 °C, k = 0.05 /min, t = 10 min, target 60 °C.
Initial difference: 90 − 20 = 70 K. Exponent: −0.05 × 10 = −0.5, and e^(−0.5) ≈ 0.60653.
Remaining difference: 70 × 0.60653 ≈ 42.457 K, so T(10) = 20 + 42.457 ≈ 62.46 °C.
Time to 60 °C: ln(70/40) / 0.05 = 0.55962 / 0.05 ≈ 11.19 min.

Each point is the calculator re-run with only “Initial temperature (°C)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Initial temperature (°C)	Temperature after t (°C)
22.5	21.51632665
45	35.16326649
67.5	48.81020634
90	62.45714618
112	76.10408602
135	89.75102587
180	117.0449056

How to read the result

The difference to ambient halves every ln 2/k minutes, here about 13.9 min. With this k the drink is near 46 °C after 20 min (one time constant) and about 36 °C after 30 min. A larger k means faster cooling: smaller or thinner objects, more surface area, stirring or wind all raise it. The temperature approaches ambient but, mathematically, never quite reaches it, so a target equal to or beyond T_a has no solution.

Common mistakes

Using a k measured in 1/s with time in minutes.
Setting the target between T₀ and T_a the wrong way round (the logarithm then fails).
Applying the model to large or poorly conducting objects with big internal gradients.
Ignoring evaporation or phase change, which add cooling not captured by the model.

Keep in mind: The exponential model is an idealisation. Radiation, changing air flow, evaporation and non-uniform internal temperature all bend real curves. Do not rely on it alone for food safety or forensic conclusions.

Frequently asked questions

How do I find k for my object?

Measure the temperature at two times and use k = ln[(T₁ − T_a)/(T₂ − T_a)] / (t₂ − t₁).

Does the law work for heating as well?

Yes. If the ambient is warmer, the same formula describes the object warming toward it.

Sources: Incropera, DeWitt, Bergman and Lavine, Fundamentals of Heat and Mass Transfer, Wiley (lumped capacitance method) · Isaac Newton, "Scala graduum caloris", Philosophical Transactions of the Royal Society, 1701

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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