Newton’s Law of Cooling

Newton's Law of Cooling describes how quickly a warm (or cold) object exchanges heat with its surroundings. The central idea is beautifully simple: the faster the temperature difference between an object and its environment, the faster the object loses (or gains) heat. A freshly brewed cup of coffee sitting in a cold room cools far more rapidly in the first few minutes than it does when it has nearly reached room temperature — this everyday observation is exactly what the law captures mathematically. Isaac Newton first stated this principle around 1701, and it applies whenever heat transfer is dominated by convection and conduction rather than radiation. In practical terms this means the law works well for moderate temperature differences (roughly less than a few hundred degrees Celsius above ambient). It is used by forensic scientists to estimate time of death, by engineers designing cooling systems for electronics, and by chefs who want to know how long to rest a roast before carving it. Mathematically, Newton's Law of Cooling is an example of an exponential decay process. The temperature gap between the object and its surroundings shrinks by the same fraction in every equal time interval — just like radioactive decay or the discharge of a capacitor. This makes the underlying differential equation both elegant and highly solvable, placing it among the most important first-order differential equations in applied science.

Key takeaways

The Proportionality Statement
Solving the Differential Equation
Understanding the Cooling Constant k
Half-Life Analogy

The Proportionality Statement

Start with the physical observation: the rate at which an object loses thermal energy is proportional to how much hotter it is than its surroundings. If T(t) is the object's temperature at time t and T_env is the (constant) environmental temperature, then the rate of change of temperature satisfies dT/dt = -k(T - T_env), where k > 0 is the cooling constant. The negative sign ensures that if T > T_env the object is cooling (dT/dt < 0), and if T < T_env it is warming (dT/dt > 0).

Solving the Differential Equation

Define the excess temperature θ(t) = T(t) - T_env. Then dθ/dt = dT/dt = -kθ. This is a standard separable ODE: dθ/θ = -k dt. Integrating both sides gives ln|θ| = -kt + C, so θ(t) = θ_0 · e^(−kt), where θ_0 = T(0) − T_env is the initial excess temperature. Substituting back, the full solution is T(t) = T_env + (T_0 − T_env)e^(−kt).

Understanding the Cooling Constant k

The constant k (units: s⁻¹ or min⁻¹) encodes everything about how efficiently heat leaves the object: the surface area exposed, the material's thermal conductivity, air circulation, and surface finish. A large k means rapid cooling — think of a thin metal spoon versus a thick ceramic mug. You can find k experimentally by measuring T at two different times and solving: k = (1/t) · ln[(T_0 − T_env)/(T(t) − T_env)].

Half-Life Analogy

Just as in radioactive decay, you can define a 'half-difference time' t_(1/2) — the time for the excess temperature to halve. Setting e^(−k·t_(1/2)) = 1/2 gives t_(1/2) = ln(2)/k ≈ 0.693/k. This is independent of the starting temperature: no matter how hot you begin, the excess always halves in the same time interval, which is a hallmark of exponential decay.

The Key Equation

T(t) = T_{\text{env}} + (T_0 - T_{\text{env}})\,e^{-kt}

T(t) — Temperature at time t (°C or K): The object's temperature at elapsed time t.
T_{\text{env}} — Environmental (ambient) temperature (°C or K): The constant temperature of the surroundings, assumed not to change.
T_0 — Initial temperature (°C or K): The object's temperature at t = 0.
k — Cooling constant (s⁻¹ (or min⁻¹)): A positive constant that depends on the object's geometry, material, and surface properties.
t — Time (s or min): Time elapsed since the start of cooling or heating.
e — Euler's number (dimensionless): The base of the natural logarithm, approximately 2.718.

Worked Example

Problem. A cup of tea is made at 95 °C and left in a room at 20 °C. After 10 minutes its temperature has dropped to 65 °C. (a) Find the cooling constant k. (b) Predict the temperature after a total of 25 minutes. (c) How long does it take to cool to a drinkable 55 °C?

k \approx 0.0511\,\text{min}^{-1},\quad T(25) \approx 40.9\,°\text{C},\quad t_{55} \approx 14.9\,\text{min}

Answer. k ≈ 0.0511 min⁻¹; temperature after 25 min ≈ 40.9 °C; reaches 55 °C after approximately 14.9 minutes.

Figure

Diagram

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · English

Newton stated his cooling law in the 1701 paper 'Scala Graduum Caloris' (A Scale of the Degrees of Heat), where he used linseed oil as a thermometric fluid and tracked how iron cooled after being removed from a furnace. By comparing cooling rates he established that the rate of temperature change was proportional to the excess temperature — the first quantitative law of heat transfer. Newton did not have a precise thermometer; he estimated temperatures partly by touch and by observing when materials like butter, wax, and tin melted — making his quantitative cooling observations all the more remarkable.

Specific Heat Capacity — The amount of heat an object stores per degree determines how quickly its temperature responds to heat flow, directly affecting the cooling constant k in Newton's Law.
Thermal Conduction — Newton's Law of Cooling applies when conduction (and convection) dominate heat transfer; understanding conduction explains why the cooling constant depends on material and geometry.
Exponential Decay & First-Order ODEs — The differential equation dT/dt = −k(T − T_env) is a first-order linear ODE whose solution is an exponential — connecting Newton's Law to a broad family of decay phenomena in maths and science.
Stefan–Boltzmann Radiation Law — At very high temperature differences, radiative heat loss (proportional to T⁴) dominates over the linear approximation, showing where Newton's Law of Cooling breaks down.