Newton's Law of Cooling Calculator

Calculate object temperature at any time during cooling or heating using Newton's law of cooling.

τ = 1/k
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Cooling Is Exponential When Heat Transfer Is Proportional to Temperature Difference

Newton’s law of cooling models the rate of temperature change as proportional to the difference between an object and its surroundings. The differential equation dT/dt=−k(T−T∞) has the solution T(t)=T∞+(T0−T∞)e−kt. The temperature difference therefore decays by the same fraction during equal time intervals.

The model works best when the surrounding temperature is approximately constant and internal temperature gradients in the object are small. The parameter k combines convection, geometry, thermal mass, and sometimes other heat-transfer effects, so it is often found experimentally for a particular object and environment.

T(t)=T∞+(T0−T∞)e−kt
SymbolMeaningWhy it appears / units
T0Initial object temperature°C or K; temperature differences have the same numerical size in either scale.
T∞Ambient temperatureConstant surrounding temperature in the basic model.
kCooling constant1/time; larger k means faster approach to ambient.
tElapsed timeMust use the time unit matching k.

The object approaches ambient temperature asymptotically and does not overshoot it in this simple first-order model. Heating toward a warmer environment follows the same equation with the sign of the initial temperature difference reversed.

The temperature difference should decay toward zero, not change sign spontaneously. In the basic model, T(t)−T∞ keeps its initial sign and shrinks exponentially. At one time constant it should retain about 36.8% of its initial magnitude.

Worked Examples

Example 1: Coffee cooling: T₀=80°C, T_∞=20°C, k=0.02/s, t=60s
T=20+(80-20)×e^(-1.2)
Result: T=20+60×0.301=38°C after 1 minute
Drink within 5 minutes for warmth
Example 2: Forensics: body found at 25°C, room 18°C, k=0.005/s
t=-ln((25-18)/(37-18))/0.005
Result: t=198 min = 3.3 hours since death
Time of death estimation
Example 3: Cooling difference halves
temperature difference halves in 10min
Result: k=ln2/10≈0.0693min−1
The half-time of the temperature difference is ln2/k.
Example 4: Heating from below ambient
T0=10°C, T∞=30°C, k=0.1min⁻¹, t=5min
Result: T≈17.9°C
The same exponential law describes heating as well as cooling.

Common Mistakes

⚠️
Assuming temperature decreases linearly with time

The rate slows as the object approaches ambient because the driving temperature difference becomes smaller.

⚠️
Using inconsistent time units for k and t

If k is per minute, t must be in minutes unless k is converted.

⚠️
Applying one constant k after conditions change

Airflow, surface conditions, phase changes, or changing surroundings can alter the effective heat-transfer rate.

Frequently Asked Questions

Newton's law of cooling validity?
Applies when temperature difference is small (linear approximation to Stefan-Boltzmann) and when convection dominates. For large ΔT or radiation-dominated cooling, more complex models are needed.
Lumped vs distributed?
Newton's cooling is 'lumped' — assumes uniform temperature in the object. Valid when Bi<0.1. For thermally thick objects, heat diffusion inside the object also matters.
Why does the curve flatten near room temperature?
The cooling rate is proportional to T−T∞. As that difference shrinks, the rate of heat transfer and temperature change also shrinks.
Can Newton’s law describe warming?
Yes. If the object begins colder than the surroundings, the same exponential solution approaches the higher ambient temperature from below.
How can k be found from data?
Measure temperature at known times with a stable ambient temperature, then fit the exponential decay of T−T∞. Two ideal data points can also determine k through a logarithm.
When is the model inaccurate?
It can fail when internal gradients are large, radiation is dominant, ambient temperature changes, heat-transfer coefficients vary strongly, or phase changes occur.
When does Newton’s law of cooling become inaccurate?
The exponential model assumes a roughly constant heat-transfer coefficient, constant ambient temperature, and a body that can be represented by one uniform temperature. Large property changes, radiation-dominated exchange, phase changes, strong spatial temperature gradients, or changing airflow can make the fitted cooling constant vary with time.

Formula Explorer connections

Interpretation: This relationship connects motion, force, momentum, work or energy in a mechanical system. Assumption: Choose a consistent reference direction and unit system. The model may assume constant acceleration, rigid bodies, negligible losses or an isolated system.

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