Newton’s Law of Cooling Calculator
Work out how fast something cools or warms to room temperature. Solve T(t) = T_env + (T₀ − T_env)e^(−kt) for the temperature at a time, the time to reach a temperature, the cooling constant k from two readings, or the starting temperature, with the cooling curve drawn.
How to Use
- Pick what to solve for: the temperature after a time, the time to reach a temperature, the cooling constant k, or the starting temperature.
- Enter the temperature of the surroundings and the other temperatures in °C, °F or K; they can be in different scales.
- Enter the time and the cooling constant k per minute, second or hour. To find k, give two readings: the first as the starting temperature and the second after the time between them.
- In the k mode, add an earlier temperature (such as 37 °C for a body) to back-date how long before the first reading it was that warm.
- Read the answer in the highlighted field, the half-life of the temperature gap, and the cooling curve heading towards the room temperature line.
- Press a preset to load an example and check Show Work for each step.
Worked Example
When will the coffee be drinkable? Coffee at 90 °C in a 21 °C room, with k = 0.035 per minute (an illustrative figure for an open mug). The gap starts at 69 °C and must fall to 60 − 21 = 39 °C, so t = ln(69 ÷ 39) ÷ 0.035 = 16.3 minutes. The gap halves every ln 2 ÷ 0.035 = 19.8 minutes.
k from two readings, then back-dating. An object reads 30 °C, then 28.5 °C an hour later, in a 20 °C room. k = ln(10 ÷ 8.5) ÷ 1 h = 0.1625 per hour. It was at 37 °C ln((37 − 20) ÷ (30 − 20)) ÷ 0.1625 = 3.27 hours before the first reading. This is the textbook time-of-death sum, shown as an illustration only.
The common mistake: using temperatures instead of gaps. For the coffee, ln(90 ÷ 60) ÷ 0.035 = 11.6 minutes, and in kelvin ln(363.15 ÷ 333.15) ÷ 0.035 = 2.5 minutes; both are wrong. The law is about the difference from the room, so the ratio must be (90 − 21) ÷ (60 − 21), giving 16.3 minutes.
Show Work
Formulas
From Newton’s Hot Iron to the Biot Number
Isaac Newton published the law anonymously in 1701, in a short paper in the Philosophical Transactions titled “Scala graduum caloris”, a scale of degrees of heat. He heated a block of iron in a fire, let it cool in a steady breeze and timed it, using the steady rate of cooling to estimate temperatures far above what his linseed-oil thermometer could read.
In 1817 Pierre Louis Dulong and Alexis Thérèse Petit showed that the law breaks down for large temperature differences, where radiation, which grows with the fourth power of absolute temperature, takes over. Joseph Fourier’s Théorie analytique de la chaleur (1822) then described how heat moves inside a body, and the ratio that tells whether a body can be treated as one temperature is named after Jean-Baptiste Biot, who studied conduction in 1804.
The law is used in forensic textbooks to back-date a time of death, but real practice uses methods such as the nomogram developed by Claus Henssge in the 1980s, which corrects for body weight, clothing, air movement and the plateau before a body starts to cool. This calculator does not implement it; the time-of-death preset is an illustration of the mathematics.
About This Tool
This calculator solves Newton’s law of cooling for whichever value you leave out: the temperature after a time, the time to reach a temperature, the cooling constant from two readings, or the starting temperature, and it can back-date two readings to an earlier temperature. Temperatures can be in Celsius, Fahrenheit or kelvin, mixed freely, and are checked against absolute zero. It reports the half-life of the temperature gap and the time constant, works for warming as well as cooling, and draws the curve approaching the room temperature. The k values in the presets are illustrative; for real objects, measure k from two readings.
Everything runs in your browser; nothing you enter is sent anywhere.
Related tools: Specific Heat Calculator, Heat Conduction Calculator, and Half-Life Calculator.
Frequently Asked Questions
What is Newton’s law of cooling?
The rate of cooling is proportional to the gap between the object and its surroundings, so the gap shrinks exponentially: T = T_env + (T₀ − T_env)e^(−kt). Coffee at 90 °C in a 21 °C room with k = 0.035 per minute (an illustrative value) is at 21 + 69 × e^(−0.525) = 61.8 °C after 15 minutes.
How do I find the cooling constant k?
Take two readings a known time apart and use k = ln((T₁ − T_env) ÷ (T₂ − T_env)) ÷ t. Water cooling from 80 °C to 50 °C in 10 minutes in a 20 °C room gives k = ln(60 ÷ 30) ÷ 10 = 0.0693 per minute: the gap halves every 10 minutes.
Can it estimate a time of death?
Only roughly, and only as an illustration. A body at 30 °C, then 28.5 °C an hour later, in a 20 °C room gives k = ln(10 ÷ 8.5) = 0.1625 per hour; assuming 37 °C at death, that was ln(17 ÷ 10) ÷ 0.1625 = 3.27 hours before the first reading. Forensic practice uses the Henssge nomogram, which allows for body weight, clothing and an early plateau; it is not implemented here.
Does it work for warming up?
Yes, the same law runs in reverse when the object starts colder. A drink at 4 °C left in a 25 °C room with k = 0.02 per minute (illustrative) is at 25 − 21 × e^(−0.6) = 13.5 °C after 30 minutes, and the gap halves every 34.7 minutes.
When does Newton’s law of cooling stop being accurate?
It assumes the object has one temperature throughout, which needs a Biot number Bi = hL ÷ k below about 0.1. With h = 10 W/(m²·K) and a characteristic length of 1 cm, that needs a thermal conductivity above 1 W/(m·K): metal parts pass easily, a human body does not. Large temperature gaps (radiation grows as T⁴), evaporation and changing surroundings also bend the curve.
How do I use the Newton’s Law of Cooling Calculator?
Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.
Is it free? Does it work without internet?
Yes to both. It is free with no sign-up, and once the page has loaded it keeps working even with no internet.
Where does my data go?
Nowhere — every calculation runs on your own device. Nothing you enter is uploaded, logged, or stored.
Common Use Cases
When is the coffee drinkable
90 °C coffee in a 21 °C room at k = 0.035/min (illustrative) takes ln(69 ÷ 39) ÷ 0.035 = 16.3 minutes to reach 60 °C.
A drink out of the fridge
A 4 °C drink in a 25 °C room with k = 0.02/min (illustrative) warms to 13.5 °C in 30 minutes.
Classroom experiment
Water at 80 °C falling to 50 °C in 10 minutes in a 20 °C room gives k = 0.0693/min and a time constant of 14.4 minutes.
How hot was it to start
Something at 40 °C after 20 minutes in a 20 °C room, with k = 0.035/min, started at 20 + 20 × e^0.7 = 60.3 °C.
Forensic textbook problems
Readings of 30 °C and 28.5 °C an hour apart at 20 °C back-date a 37 °C body by 3.27 hours: an illustration, not a forensic method.
Last updated: