Newton's Law of Cooling Calculator
Estimate how quickly an object cools down or warms up to the ambient temperature. Enter the initial and ambient temperatures, the cooling constant k, and the elapsed time to get the object's temperature, plus the time needed to reach any target temperature.
Predict Cooling or Warming of an Object over Time
Newton's Law of Cooling Formula
Where:
- \(T(t)\) — temperature of the object after time \(t\) (°C, °F or K)
- \(T_0\) — initial temperature of the object at \(t = 0\)
- \(T_a\) — ambient temperature of the surroundings (assumed constant)
- \(k\) — cooling constant, a positive number describing how fast heat is exchanged (per second, minute or hour)
- \(t\) — elapsed time, in units consistent with \(k\)
The formula says that the object's temperature approaches the ambient temperature exponentially: the difference \(T - T_a\) shrinks by the same factor \(e^{-k}\) every unit of time, regardless of how hot or cold the object is.
If you instead want the time needed to reach a given target temperature \(T_t\), invert the formula:
\[ t = -\frac{\ln\left(\dfrac{T_t - T_a}{T_0 - T_a}\right)}{k} \]
Newton's Law of Cooling – Calculation Example
Example: a cup of coffee at 90 °C is left in a room at 20 °C. The cooling constant is about \(k = 0.05\) per minute. What is its temperature after 10 minutes, and how long until it cools to 40 °C?
\[ T(10) = 20 + (90 - 20)\,e^{-0.05 \times 10} = 20 + 70\,e^{-0.5} \approx 20 + 70 \times 0.6065 \approx 62.46\ \text{°C} \]
\[ t = -\frac{\ln\left(\dfrac{40 - 20}{90 - 20}\right)}{0.05} = -\frac{\ln(0.2857)}{0.05} \approx \frac{1.2528}{0.05} \approx 25.06\ \text{min} \]
Result: after 10 minutes the coffee is at about 62.46 °C, and it reaches 40 °C after roughly 25 minutes (about 25.06 min).
How to use the calculator
- Enter the Initial temperature (T₀) of the object and pick its unit (°C, °F or K).
- Enter the Ambient temperature (Tₐ) — the constant temperature of the surroundings — and its unit.
- Enter the Cooling constant (k) and choose whether it is given per second, per minute or per hour. It must be greater than zero.
- Enter the Elapsed time (t) since the start, with its unit.
- Optionally, enter a Target temperature. If you do, the calculator also reports how long it takes for the object to reach that temperature. Leave this field blank to skip it.
- Read the results: the object's temperature after the elapsed time (shown in °C, °F and K) and, if a target was set, the time needed to reach it.
What is Newton's law of cooling?
Newton's law of cooling states that the rate at which an object loses (or gains) heat is proportional to the temperature difference between the object and its surroundings. In other words, the hotter the object is relative to the room, the faster it cools; as it gets closer to the ambient temperature, the cooling slows down. Solving the resulting differential equation gives the exponential formula used above: the temperature difference decays steadily, shrinking by the same percentage each time interval.
A key consequence is that an object never exactly reaches the ambient temperature — it only approaches it asymptotically. In practice, the difference quickly becomes so small that it is undetectable. That is why the calculator warns you if you set the target temperature equal to the ambient temperature: mathematically, it takes infinite time. It is better to aim for a value slightly above or below the ambient one, such as "within 1 °C of room temperature".
The same law works for warming. If the initial temperature is below the ambient temperature (a cold drink in a warm room, for instance), the formula still applies — the temperature difference is simply negative, and the object warms up exponentially toward the surroundings. The calculator handles both directions automatically.
The law is used in engineering (heat exchangers, quenching of metal parts), forensic science (estimating time of death from body cooling), food safety (cooling of cooked food) and everyday situations like tea or coffee going cold. It is the same exponential-decay mathematics that appears in radioactive decay and capacitor discharge.
Choosing the cooling constant k
The cooling constant \(k\) depends on the object's size, shape, material and the surrounding medium — not just on the object itself. The same object cools much faster in stirred water than in still air. Some typical orders of magnitude:
- A cup of coffee or tea in still air: \(k \approx 0.02\)–\(0.08\) per minute
- A small metal part in a water bath: \(k\) can reach several per minute
- A large roast taken out of the oven: \(k \approx 0.01\)–\(0.03\) per minute
If you don't know \(k\), you can measure it: note the temperature at two known times, plug the values into the formula, and solve for \(k\). Alternatively, run this calculator with a guessed \(k\), compare the predicted temperature at your measurement time, and adjust until it matches.
Assumptions and limitations
- Constant ambient temperature. The surroundings must stay at \(T_a\) throughout — a large room or a well-stirred bath. A small closed space heats up along with the object.
- Uniform object temperature. The object is assumed to have the same temperature throughout (lumped-system assumption). This is a good approximation for small objects made of good thermal conductors; it fails for large or poorly conducting bodies, which cool from the outside in.
- Linear heat transfer. The law models convective heat transfer with a constant heat-transfer coefficient. It ignores radiation (significant for very hot objects) and phase changes such as boiling or freezing.
- Exponential behavior. Because cooling is exponential, the last few degrees near the ambient temperature take disproportionately long — doubling the temperature difference always takes the same time, no matter how small the difference already is.
For related physics calculations, see our Free Fall Calculator, the Drag Force Calculator (another situation where a velocity-like quantity decays exponentially in time), or the Speed of Sound in Air Calculator for other temperature-dependent properties of air.
Frequently Asked Questions
How do I calculate Newton's law of cooling by hand?
Plug your values into \( T(t) = T_a + (T_0 - T_a)\,e^{-kt} \) step by step. First find the initial temperature difference \(T_0 - T_a\), then multiply it by \(e^{-kt}\) (use the \(e^x\) key on a scientific calculator), and finally add \(T_a\) back. For example, with \(T_0 = 90\), \(T_a = 20\), \(k = 0.05\) per minute and \(t = 10\): \(20 + 70 \times e^{-0.5} \approx 62.5\) °C.
How can I find the cooling constant k if I don't know it?
Measure the object's temperature at two known times, then solve for \(k\): \( k = -\frac{1}{t}\ln\left(\frac{T(t) - T_a}{T_0 - T_a}\right) \). For instance, if a drink drops from 80 °C to 50 °C in 15 minutes in a 20 °C room, \(k \approx 0.052\) per minute. One measurement pair is enough in theory, but averaging several improves accuracy, since real cooling data is never perfectly exponential.
Can I use Fahrenheit or Kelvin temperatures?
Yes — the mathematics works in any consistent temperature scale because only temperature differences appear in the exponential term. Just make sure \(T_0\), \(T_a\) and the target temperature all use the same unit. The calculator converts °F and K to °C internally and shows the result in all three units, so you don't need to convert anything yourself.
How long does it take to get within 1 degree of ambient temperature?
Use \( t = -\ln\left(\frac{1}{T_0 - T_a}\right)/k \) with the small remaining difference (e.g. 1 °C) in the numerator. Note that reaching exactly \(T_a\) is mathematically impossible — the object only approaches it asymptotically, which is why the last degree takes longer than the first ten. If you need a practical answer, always aim for "within X degrees" rather than the ambient value itself.
What are the most common mistakes when using this formula?
The biggest one is mixing units between \(k\) and \(t\): if \(k\) is per minute, \(t\) must be in minutes, not seconds or hours. Other frequent errors are using a negative \(k\) (it must be positive, with the minus sign already built into the formula), forgetting to subtract \(T_a\) before exponentiating, and assuming the room temperature stays constant when the object actually warms or cools its surroundings.
Is Newton's law of cooling accurate for estimating time of death?
It is a useful first approximation in forensic science, but real body cooling deviates from the simple exponential model. There is an early "plateau" phase while internal heat equalizes, and the cooling constant varies with body size, clothing, air movement and water immersion. Professional methods apply corrective rules on top of the basic law. Treat any result from this calculator as a rough estimate, not evidence.
Is there a half-life for cooling, like in radioactive decay?
Yes. The time to halve the temperature difference is constant and equals \( t_{1/2} = \ln 2 / k \approx 0.693/k \), independent of how hot the object is. A cup with \(k = 0.05\) per minute halves its gap to room temperature every ~13.9 minutes: 70 °C above ambient becomes 35 °C, then 17.5 °C, and so on. This is a handy sanity check on your results.