Newton law of cooling calculator
This Newton law of cooling calculator solves T(t) = Ts + (T0 − Ts) · e^(−κt): a 90 °C drink in a 20 °C room with κ = 0.05 per minute is at 62.46 °C after 10 minutes.
Example: Calculate: Temperature after a time · Temperature unit: °C · Time unit: minutes · Start temperature T0 90 · Surrounding temperature Ts 20 · Cooling constant κ 0.05 · Elapsed time t 10 → Result: After 10 min: 62.46 °C. Source: OpenStax Calculus Volume 2, section 4.3: Newton’s law of cooling dT/dt = k(T − Ts), solution T(t) = Ts + (T0 − Ts)e^(kt); pizza example Ts = 75 °F, T0 = 200 °F, k ≈ −0.08338, 150 °F after about 6.12 min (read 13.10.2026). Updated: .
How it is calculated
Newton’s law of cooling: T(t) = Ts + (T0 − Ts) · e^(−κ·t). T0 is the start temperature, Ts the temperature of the surroundings, t the time and κ the cooling constant per time unit. The law says that the temperature changes at a rate proportional to the difference from the surroundings (OpenStax Calculus Volume 2, with k = −κ).
Example: a hot drink
T0 = 90 °C, Ts = 20 °C, κ = 0.05 per minute. After 10 minutes: T = 20 + 70 · e−0.5 ≈ 62.46 °C. The drink reaches 60 °C after ln(70 ÷ 40) ÷ 0.05 ≈ 11.19 minutes, and the difference to the room halves every ln 2 ÷ 0.05 ≈ 13.86 minutes.
| Minutes | Temperature in °C |
|---|---|
| 0 | 90.00 |
| 5 | 74.52 |
| 10 | 62.46 |
| 20 | 45.75 |
| 30 | 35.62 |
| 60 | 23.49 |
Worked example from OpenStax
A pizza at 200 °F in a 75 °F kitchen cools to 190 °F in one minute. Then κ = ln(125 ÷ 115) ≈ 0.08338 per minute, so it is at 172.34 °F after 3 minutes and reaches 150 °F after about 6.13 minutes (the book prints 6.12 because it rounds k first).
The same exponential shape describes decay, see the half-life calculator; for the final temperature of mixed liquids use the mixing temperature calculator.
Limits
The law fits when the surroundings stay constant and heat leaves mainly by conduction or convection. κ must be measured for your object, for example with the third mode.
Frequently asked questions
What is Newton’s law of cooling?
The temperature of an object changes at a rate proportional to its difference from the surroundings, giving T(t) = Ts + (T0 − Ts) · e^(−κt).
How do I find the cooling constant?
Measure the temperature T1 after a time t1: κ = ln((T0 − Ts) ÷ (T1 − Ts)) ÷ t1. For 90 → 80 °C in 5 min at 20 °C that is 0.03083 per minute.
How long until a hot drink is at a given temperature?
t = ln((T0 − Ts) ÷ (T − Ts)) ÷ κ. With T0 = 90, Ts = 20, κ = 0.05 and T = 60 that is about 11.19 minutes.
Does it work in Fahrenheit?
Yes, the formula only uses temperature differences; choose °F for the display and keep all temperatures in the same unit.
Why does it never reach room temperature exactly?
The difference shrinks by the same factor in equal time steps, so it approaches zero but never reaches it in theory.
Sources and legal basis
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