Newton's law of cooling

Take two identical cups of tea, one poured at $90\,^\circ\text{C}$ and left on the same kitchen table at room temperature. In the first minute it loses a startling amount of heat and the surface stops steaming quite so hard; an hour later, cooled to nearly room temperature, it barely changes at all even though it is still, technically, cooling. Something about how far above the room the tea is seems to control how fast it loses heat — and that single observation, first written down by Isaac Newton in the 1700s, is the whole of Newton's law of cooling. The idea sounds almost too simple to be physics: the hotter something is compared with its surroundings, the faster it cools; the closer it gets to room temperature, the more sluggish the cooling becomes. But that simplicity is exactly what makes it a workhorse in a Class 11 classroom, a forensic lab estimating a time of death, and a kitchen thermometer app alike. It explains why a bowl of soup is scalding for the first minute and then stays "just warm" for what feels like forever, why a kettle left off the stove never quite reaches the coldness of the room in any hurry, and why cooling curves are gentle sweeping bends rather than straight ramps down to room temperature. This article builds that picture from the underlying proportionality, shows exactly how CBSE expects you to turn it into a usable numerical formula (the average-temperature method — not a full calculus solution, which sits outside the Class 11 syllabus), and works through the same beaker-of-water numbers your NCERT-aligned practical uses, extended into a second stage of cooling and a tougher multi-stage problem. By the end you should be able to state the law precisely, sketch its curve, and turn cooling data into a working rate constant $k$ with confidence.

Key takeaways

Newton's law of cooling states that a body's rate of loss of heat is proportional to the excess temperature over its surroundings, $\dfrac{dQ}{dt} \propto (T - T_s)$ , provided that excess is small.
Because $Q = mcT$ , the rate of fall of temperature is also proportional to the excess: $\dfrac{dT}{dt} \propto -(T-T_s)$ .
The cooling curve (temperature vs time) is steep at first and flattens as $T$ approaches $T_s$ — it never becomes a straight line.
CBSE Class 11 numericals use the practical average-excess-temperature version, $\dfrac{\theta_1-\theta_2}{t} = k\left(\dfrac{\theta_1+\theta_2}{2}-\theta_0\right)$ , not the exponential solution of the differential equation.
The constant $k$ depends on the body's surface area, mass, specific heat, and nature of the surface, and is treated as fixed for one body in one set of surroundings over small temperature ranges.
Because equal temperature drops take progressively longer as a body approaches room temperature, later stages of cooling always take more time than earlier, equal-sized stages.
Key idea: a body loses heat at a rate proportional to how much hotter it is than its surroundings, so equal temperature drops take longer and longer as it approaches room temperature — captured numerically by the average-excess-temperature formula.

The core insight

The rate at which a body loses heat (and hence cools) is proportional to the excess of its temperature over the surroundings, so cooling is fast when the excess is large and slows continuously as the body approaches room temperature.

Where it fits in the syllabus

Newton's law of cooling belongs to the CBSE Class 11 Physics unit Thermal Properties of Matter, right after heat transfer by conduction, convection, and radiation — it is really a radiation-and-convection consequence dressed up as its own named law. The CBSE 2025-26 prescription is explicit about how much of it you need: the qualitative statement of the law and its practical, average-temperature numerical form are examinable, but deriving the full exponential-decay solution of the underlying differential equation (the calculus route you may see in a lab manual or a higher-level course) is not part of the Class 11 theory syllabus. This article stays faithful to that boundary — every equation here is algebra, not calculus. A Class 11 student is expected to state the law in words, recognise the shape of a cooling curve, and use the average-excess-temperature formula to solve two-stage or multi-stage cooling problems, always remembering that the law only holds for small temperature differences from the surroundings.

The physical picture

A hot object loses energy to its cooler surroundings mainly by radiation, with convection currents in the surrounding air often helping the process along. Newton's insight, tested experimentally long before anyone had the mathematics of radiation laws, was that when the temperature excess is modest, the rate of heat loss tracks that excess in the simplest possible way — directly, not through any higher power.

\frac{dQ}{dt} \propto (T - T_s)

where: $\dfrac{dQ}{dt}$ is the rate at which the body loses heat energy (its instantaneous cooling power), $T$ is the body's own instantaneous temperature, and $T_s$ is the constant temperature of the surroundings. Read physically, this says: double the excess temperature and you double the rate of heat loss; let the excess shrink to zero and the heat loss shrinks to zero with it. Since heat lost is tied to temperature fall through $Q = mcT$ for a body of mass $m$ and specific heat capacity $c$ , the same proportionality carries straight over to the temperature itself:

-\frac{dT}{dt} \propto (T - T_s)

where: the minus sign simply records that $T$ is decreasing while $(T-T_s)$ is positive (the body is hotter than the room). This single line is the entire content of Newton's law of cooling — everything else in this article is either turning it into a usable exam formula or working through its consequences.

Why the rate keeps slowing down. Early on, with the body freshly poured or freshly removed from a flame, $(T-T_s)$ is large and the cooling is brisk. As $T$ drifts down toward $T_s$ , the excess shrinks, and the rate of cooling shrinks in step — the body never actually "runs out" of heat to lose in a fixed way, it simply loses it more and more gently. This is why a real cooling curve is steep at the start and bends over into a long, flattening tail as it creeps toward room temperature, never quite becoming a horizontal line within any finite time you would sit and watch.

Figure

A temperature versus time cooling curve starting steep near ninety degrees and flattening as it approaches room temperature at twenty degrees, with a shaded excess temperature gap shown at two points to show the gap shrinking over time

Turning the proportionality into a number. For a Class 11 numerical, the proportionality is converted into an equation using a fixed constant $k$ that depends on the body's surface, area, mass and specific heat: $-\dfrac{dT}{dt} = k(T-T_s)$ . Rather than solving this differential equation with calculus (which would give an exponential decay, outside the prescribed syllabus), CBSE numericals use a simpler, purely algebraic trick — approximate both sides over a short time interval using average values, exactly as derived below.

Derivation AThe average-excess-temperature cooling formula

The goal is to turn the proportionality $-\dfrac{dT}{dt} \propto (T-T_s)$ into a formula that can be solved with ordinary algebra for a body cooling from a known temperature $\theta_1$ to a known temperature $\theta_2$ in a measured time $t$ , in surroundings held at $\theta_0$ — without integrating the differential equation, since the exponential solution lies outside the Class 11 prescription. The trick is to replace the instantaneous rate and the instantaneous excess temperature, both of which change continuously during the cooling, with their average values over the interval, which is an excellent approximation whenever the interval is short and the drop is modest compared with the excess over the room.

Step 1. Write the law with a proportionality constant. Since $-\dfrac{dT}{dt}$ is proportional to $(T-T_s)$ , introduce a constant $k$ (fixed for a given body, given surface, and given surroundings, over small excess temperatures) so that $-\dfrac{dT}{dt} = k(T-T_s)$ . This constant absorbs everything about the body's geometry and material that Newton's proportionality does not specify on its own.

Step 2. Replace the instantaneous rate on the left with an average rate over the measured interval. If the body cools from $\theta_1$ to $\theta_2$ in time $t$ , the average rate of fall of temperature over that interval is simply the total drop divided by the total time, $\dfrac{\theta_1-\theta_2}{t}$ . This is not an approximation of calculus notation — it is the definition of an average rate, and it is exactly what a stopwatch-and-thermometer experiment actually measures.

Step 3. Replace the instantaneous excess temperature on the right with its average over the same interval. Because $(T-T_s)$ changes smoothly and (for a small enough temperature range) roughly evenly with time, its average value over the interval is well approximated by the excess evaluated at the average of the two end temperatures: $\left(\dfrac{\theta_1+\theta_2}{2}\right) - \theta_0$ . This is the one approximating step in the whole derivation, and it is exactly why the law is only trustworthy for small temperature differences — the flatter the curve's bend across the interval, the better this average stands in for every instantaneous value inside it.

Step 4. Equate the two averaged sides. Substituting both averages into $-\dfrac{dT}{dt}=k(T-T_s)$ gives the working formula every CBSE numerical is built on:

\frac{\theta_1-\theta_2}{t} = k\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right)

Here $k$ carries units of inverse time (per minute, per second, whatever the problem uses), and this single equation is enough to find $k$ from one stage of cooling, or to predict the time for a further stage once $k$ is known — which is exactly how both worked examples below are built.

Worked examples

Board level - finding the cooling constant from a beaker of water

A beaker of hot water cools from $80\,^\circ\text{C}$ to $70\,^\circ\text{C}$ in $2$ minutes, in a room kept at $20\,^\circ\text{C}$ . Find the cooling constant $k$ , in units of per minute.

Step 1. Identify the given values against the average-excess formula: $\theta_1 = 80\,^\circ\text{C}$ , $\theta_2 = 70\,^\circ\text{C}$ , $t = 2\ \text{min}$ , and $\theta_0 = 20\,^\circ\text{C}$ .

Step 2. Compute the average rate of fall on the left side: $\dfrac{\theta_1-\theta_2}{t} = \dfrac{80-70}{2} = \dfrac{10}{2} = 5\,^\circ\text{C per minute}$ .

Step 3. Compute the average excess temperature on the right side: $\dfrac{\theta_1+\theta_2}{2} - \theta_0 = \dfrac{80+70}{2} - 20 = 75-20 = 55\,^\circ\text{C}$ .

Step 4. Solve for $k$ : $5 = k\times 55$ , so $k = \dfrac{5}{55} = 0.0909\ \text{min}^{-1}$ . The cooling constant for this beaker, in this room, is $k \approx 0.091$ per minute, and because $k$ depends only on the beaker's surface, mass and specific heat (not on which stage of cooling you happen to be watching), it can be reused to predict any later stage of the same beaker's cooling in the same room.

Competitive level - predicting a second and a third stage of cooling

Using the same beaker and $k = 0.0909\ \text{min}^{-1}$ found above, estimate (a) the time for the water to cool further from $70\,^\circ\text{C}$ to $60\,^\circ\text{C}$ , and (b) the time for a third stage from $60\,^\circ\text{C}$ to $50\,^\circ\text{C}$ , with the room still at $20\,^\circ\text{C}$ . Compare all three stages and check that the pattern matches what Newton's law predicts.

Step 1. Stage two — average excess temperature: $\dfrac{70+60}{2}-20 = 65-20 = 45\,^\circ\text{C}$ . Apply the formula: $\dfrac{70-60}{t_2} = 0.0909\times 45 = 4.09$ , so $t_2 = \dfrac{10}{4.09} \approx 2.44\ \text{minutes}$ .

Step 2. Stage three — average excess temperature: $\dfrac{60+50}{2}-20 = 55-20 = 35\,^\circ\text{C}$ . Apply the formula again: $\dfrac{60-50}{t_3} = 0.0909\times 35 = 3.18$ , so $t_3 = \dfrac{10}{3.18}\approx 3.14\ \text{minutes}$ .

Step 3. Compare the three equal $10\,^\circ\text{C}$ drops: $t_1 = 2.00\ \text{min}$ , $t_2 \approx 2.44\ \text{min}$ , $t_3\approx 3.14\ \text{min}$ . Each successive stage takes noticeably longer than the last, even though the temperature drop is the same $10\,^\circ\text{C}$ every time. Answer: the second stage takes about $2.44$ minutes and the third about $3.14$ minutes, confirming that equal temperature drops take progressively longer as the beaker approaches room temperature. Timing note: under two minutes per stage once the average-excess formula is set up. The whole calculation is two divisions and one multiplication per stage — the only trap is remembering to recompute a fresh average excess temperature for every new pair of end-points; reusing an earlier stage's average excess with a later stage's temperature drop is the single most common slip under time pressure.

Common traps

Forgetting the "small temperature difference" condition. The averaging step in the derivation above only works well when the excess temperature does not change too much across the interval. For very hot bodies (molten metal, a naked flame) held far above the surroundings, heat loss is governed by Stefan's law, $E=\sigma T^4$ , which is not linear in the excess, and Newton's law becomes a poor approximation.
Using $\theta_1$ or $\theta_2$ alone instead of their average. The formula needs the average of the two end temperatures, $\dfrac{\theta_1+\theta_2}{2}$ , not either endpoint by itself — plugging in just $\theta_1-\theta_0$ or just $\theta_2-\theta_0$ is a very common exam slip.
Mixing up the units of $k$ . A value of $k$ found in per-minute must be converted before it is reused with a time given in seconds, and vice versa — $k$ is only a pure number once its time unit is fixed.
Assuming the cooling curve is a straight line. Because the rate of cooling depends on the current excess temperature, the temperature-versus-time graph bends and flattens continuously; it is never a straight-line graph, and sketching it as one loses marks even when the numerical answer is right.
Forgetting that later, equal-sized drops take longer. As the worked competitive example shows, a $10\,^\circ\text{C}$ drop near the start is quicker than the same $10\,^\circ\text{C}$ drop later — treating cooling as a constant rate throughout is the conceptual root of most numerical errors here.

Common trapNewton's law of cooling is not the exponential-decay formula, and it only holds for small excess temperatures

Two errors dominate exam scripts on this topic. The first is quoting or attempting to derive the exponential solution $T = T_s + (T_0-T_s)e^{-kt}$ as if it were the required CBSE Class 11 answer — that calculus-based solution belongs to a higher level and the practical/lab treatment, not the Class 11 theory syllabus, where the expected tool is the algebraic average-excess-temperature formula derived above. The second is applying Newton's law unconditionally to any hot object, including ones far above room temperature, such as molten metal or a flame. The law is a good approximation only when the excess temperature is small; for large excess temperatures the actual heat-loss law is Stefan's radiation law, $E=\sigma T^4$ , which grows far faster than a simple proportionality and gives a visibly different (much steeper, non-linear) cooling behaviour.

You should be able to…

State Newton's law of cooling in your own words, including the condition that it holds only for small excess temperatures over the surroundings.
Explain, physically, why a hot object cools faster at the start and progressively more slowly as it nears room temperature.
Sketch, or recognise, the general steep-then-flattening shape of a temperature-versus-time cooling curve, and explain why it is never a straight line.
Derive and apply the average-excess-temperature formula to find $k$ from one stage of measured cooling data.
Use a previously found value of $k$ to predict the time for a further stage, or a completely different pair of temperatures, for the same body in the same surroundings.
Explain why Newton's law breaks down for very large excess temperatures, and name Stefan's law as the correct alternative in that regime.

Ask the tutor: Want to try a fresh multi-stage cooling problem with your own starting temperature, or see why the averaging step stops being accurate for a large temperature drop — ask the AI tutor to work through it with you.

Why do we use the average of the two temperatures instead of just the starting temperature?

Because the rate of cooling is not constant during the interval — it keeps changing as the excess temperature shrinks. Using only the starting excess would overestimate the true rate for the whole interval (since the body cools more slowly than that once it has partly cooled), while using only the final excess would underestimate it. Splitting the difference with the average of the two end temperatures gives a much closer match to the true, continuously changing rate, and it is simple enough to compute with ordinary algebra rather than calculus — which is exactly why CBSE Class 11 uses this average-temperature version of the law.

Is Newton's law of cooling the same as the exponential decay formula I have seen online?

Not for Class 11 purposes. The full mathematical treatment does lead to an exponential decay curve, $T = T_s + (T_0-T_s)e^{-kt}$ , if you solve the underlying differential equation with calculus — but that derivation sits above the current CBSE Class 11 prescription and belongs to the lab/practical or a later course. What you are expected to know and use is the algebraic average-excess-temperature formula worked through in this article, which gives essentially the same physical picture (fast cooling early, slow cooling late) without needing to solve a differential equation.

Does Newton's law of cooling work for something as hot as a red-hot iron rod?

Not accurately, no. The law assumes the excess temperature over the surroundings is small enough that heat loss scales in simple direct proportion to that excess. A red-hot iron rod is hundreds of degrees above room temperature, and at that scale the dominant heat-loss mechanism is radiation following Stefan's law, $E = \sigma T^4$ — which rises far more steeply with temperature than a simple proportionality. Newton's law is the right tool for a warm cup of tea or a beaker a few tens of degrees above room temperature, not for glowing metal.

Why does the cooling curve flatten out instead of the object just reaching room temperature and stopping?

Because the rate of cooling itself depends on how far above room temperature the object still is. As the object gets closer to $T_s$ , the excess temperature driving the heat loss keeps shrinking, so the cooling keeps getting gentler and gentler — it approaches room temperature more and more slowly rather than hitting it and stopping abruptly. In principle (in the idealised exponential picture beyond this syllabus) the object only reaches $T_s$ after an infinite time; in practice, other effects eventually take over and the object settles at room temperature for all practical purposes.