Newton's law of cooling

A cup of tea at $80^\circ\text{C}$ left in a $30^\circ\text{C}$ room does not cool at a steady rate. It plunges fast in the first few minutes, then dawdles as it nears room temperature, and mathematically never quite gets there. Newton's law of cooling captures this everyday fact: the rate of heat loss depends on how far above the surroundings a body is, not on its temperature in isolation. Twice the temperature excess means twice the cooling rate. This is the closing topic of the Heat transfer section in Unit 8, "Heat and Thermodynamics," coming right after conduction, convection, and the radiation laws. The syllabus is explicit that deriving the law from Stefan's law is not required — you need the statement, the graph, and the ability to crunch numericals, in both the instantaneous and the average-temperature forms. Below, the core proportionality is set up, its most exam-useful average-interval form is derived, and both are applied to worked numbers so you can see exactly how a cooling curve flattens as it approaches room temperature.

The core insight

The rate of cooling of a body is directly proportional to the excess of its temperature over the surroundings, $-\frac{dT}{dt} = k(T-T_s)$ , so cooling is fast when hot and slows to a crawl as $T \to T_s$ .

The core idea

The statement. A body loses heat — and its temperature falls — at a rate proportional to the temperature difference between the body and its surroundings, provided that difference stays small.
Why "small" matters. Real heat loss (conduction, convection, radiation combined) is a more complicated function of temperature; Newton's law is the linear approximation that holds only when the body is not drastically hotter than its surroundings.
The shape of the curve. Plotted against time, temperature falls steeply at first and flattens as it approaches $T_s$ , giving an exponential-decay-like curve that never crosses below $T_s$ .
Where it sits among the other heat-transfer laws. Conduction and convection carry heat through a medium in contact with the body, while radiation obeys Stefan's law with a $T^4$ dependence; Newton's law of cooling is not a separate physical mechanism but a convenient linear stand-in for the combined effect of all three, valid only when the body is not far above room temperature.

The instantaneous form is:

-\frac{dT}{dt} = k(T - T_s)

where: $T$ is the instantaneous temperature of the body, $T_s$ is the constant temperature of the surroundings, $t$ is time, and $k$ is a positive cooling constant fixed by the body's surface area, material, and heat capacity. The minus sign on the left just records that $T$ decreases with time whenever $T > T_s$ ; the law does not fire in reverse to predict overshoot below $T_s$ . Because an instantaneous rate is awkward to measure directly, the exam-friendly version replaces it with an average rate over a finite interval, and swaps the instantaneous $T$ for the average of the initial and final temperatures:

\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_s\right)

where: $T_1$ and $T_2$ are the temperatures at the start and end of an interval of duration $t$ , and $\frac{T_1+T_2}{2}$ stands in for "the temperature during the interval." This approximation is only good when $T_1 - T_2$ is small, which is exactly the regime the law is meant for.

Figure

Cooling curve of temperature against time starting steep near eighty degrees and flattening as it approaches a dashed surroundings line at thirty degrees

Derivation AThe average-interval form of Newton's law

Starting from the instantaneous proportionality $-\frac{dT}{dt} = k(T-T_s)$ , show why replacing the exact rate over a short cooling interval with the average rate $\frac{T_1-T_2}{t}$ and the instantaneous temperature with the average $\frac{T_1+T_2}{2}$ gives an equation that is accurate enough for numericals whenever the temperature drop in that interval is small.

Step 1. Treat the interval from $T_1$ to $T_2$ (occurring over time $t$ ) as short enough that $k$ and $T_s$ can be taken as constant throughout — this is guaranteed by the same "small temperature difference" condition the law already requires.

Step 2. Over this short interval, the average rate of temperature fall is simply the total drop divided by the total time, $\frac{T_1-T_2}{t}$ , which approximates $-\frac{dT}{dt}$ evaluated somewhere inside the interval rather than at a single instant.

Step 3. Since $k(T-T_s)$ varies only slightly across the interval (because $T_1-T_2$ is small), its value at the midpoint temperature is a good stand-in for its average value across the whole interval — so $k(T-T_s)$ is replaced by $k\left(\frac{T_1+T_2}{2}-T_s\right)$ .

Step 4. Equating the two approximations gives the working formula $\frac{T_1-T_2}{t} = k\left(\frac{T_1+T_2}{2}-T_s\right)$ , which lets you compute $k$ from one measured interval and then predict $t$ or a temperature for a second interval, exactly as required by the syllabus.

Worked example

A liquid cools from $80^\circ\text{C}$ to $70^\circ\text{C}$ in 5 minutes in surroundings at $30^\circ\text{C}$ . Find the time taken to cool further from $70^\circ\text{C}$ to $60^\circ\text{C}$ .

Step 1. For the first interval, $T_1 = 80$ , $T_2 = 70$ , $t_1 = 5\ \text{min}$ , and the average temperature is $\frac{80+70}{2} = 75^\circ\text{C}$ .

Step 2. Substitute into the average-interval form to find $k$ :

\frac{80-70}{5} = k(75-30) \implies 2 = 45k \implies k = \frac{2}{45}\ \text{min}^{-1}

Step 3. For the second interval, $T_1 = 70$ , $T_2 = 60$ , average temperature $= 65^\circ\text{C}$ , and the same $k$ applies since the liquid and surroundings are unchanged:

\frac{10}{t_2} = \frac{2}{45}(65-30) = \frac{2}{45}(35) = \frac{70}{45} \approx 1.556\ ^\circ\text{C/min}

Step 4. Solve for $t_2$ :

t_2 = \frac{10}{1.556} \approx \mathbf{6.43\ minutes}

The second 10-degree drop takes noticeably longer than the first (6.43 min versus 5 min) — direct confirmation that cooling slows as the liquid approaches room temperature.

Competitive stretch

Using the same $k = \frac{2}{45}\ \text{min}^{-1}$ , estimate the instantaneous rate of cooling right at $t=0$ , when the liquid is at $80^\circ\text{C}$ . Using $-\frac{dT}{dt} = k(T-T_s) = \frac{2}{45}(80-30) = \frac{2}{45}(50) \approx 2.22\ ^\circ\text{C/min}$ . Compare this to the average rate found over the first interval, $\frac{10}{5}=2\ ^\circ\text{C/min}$ — the instantaneous rate at the start of the interval is slightly higher than the interval average, exactly as expected since the rate is falling throughout the interval. This also shows why the average-interval formula is only an approximation: it quietly smooths out the fact that the true rate keeps changing even within a single 5-minute window, which is fine as long as the temperature drop within that window stays small.

Common traps

Applying the law when the temperature excess is large — it is only a good approximation for small $(T-T_s)$ ; for big differences, the true $T^4$ radiative dependence takes over and the linear law fails.
Using $T_1$ (the starting temperature) instead of the average $\frac{T_1+T_2}{2}$ in the interval form — this is the single most common numerical slip.
Assuming $T_s$ can change during the process — the law only holds for a fixed surrounding temperature.
Treating $k$ as a universal constant — it depends on the specific body's shape, surface, and material, and must be found from given data, not recalled from memory.
Mixing up which temperature is which in the interval formula — $T_1$ is always the earlier (hotter) reading and $T_2$ the later (cooler) one, so $T_1-T_2$ and the average $\frac{T_1+T_2}{2}$ must be computed in that order every time.

Ask the tutor: Stuck on a two-interval cooling problem or unsure which temperature to average — walk through your own numbers with the AI tutor.

Why does the tea cool fast at first and slowly later, even though no heat is being added

Because the cooling rate itself depends on the temperature excess $(T-T_s)$ , not on some fixed rate of heat loss. When the tea is $50^\circ\text{C}$ above room temperature, it loses heat quickly; as it approaches room temperature, that excess shrinks, so the loss rate shrinks too. It is the same physical process throughout — only the driving temperature difference is changing, which is exactly what makes the curve steep-then-flat rather than a straight line.

Do I need to derive Newtons law of cooling from Stefans law for the ISC exam

No. The syllabus explicitly says this derivation is not required. You are expected to know the statement of the law, both its instantaneous and average-interval mathematical forms, be able to sketch and read a cooling curve, and solve numericals using the interval form — the connection to the $T^4$ radiation law is background context, not examinable derivation.

Why do we use the average of T1 and T2 instead of just T1 in numerical problems

Because the interval form of the law is an approximation to the instantaneous law over a finite stretch of time, and using the starting temperature $T_1$ throughout would overestimate the cooling rate for the whole interval, since the body is actually cooler than $T_1$ for most of that time. The average $\frac{T_1+T_2}{2}$ better represents the "typical" temperature during the interval, which is why it appears in the syllabus formula and in every mark scheme.

Newton's law of cooling

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads