Newton's Law of Cooling

In short

Newton's law of cooling says that when a body is a little warmer than its surroundings, the rate at which it loses heat is proportional to the temperature difference. In symbols,

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

where T is the body's temperature, T_s is the (constant) surrounding temperature, and k > 0 is a cooling constant that depends on the body's surface area, material, and how fast air can whisk heat away from it. Separating variables and integrating gives the solution

\boxed{\;T(t) = T_s + (T_0 - T_s)\,e^{-kt}\;}

a clean exponential decay from the initial temperature T_0 toward the ambient T_s. Taking logs of (T - T_s) linearises the curve: plotting \ln(T - T_s) against t yields a straight line of slope -k, which is how the law is verified in a school lab.

The law is an approximation. It is really Stefan's fourth-power radiation law flattened to first order for small \Delta T = T - T_s, combined with the linear contribution from convection. It works well for temperature differences up to about 30 K and fails badly when T and T_s differ by hundreds of kelvins (a red-hot iron radiates much faster than the linear law predicts).

Applications are everywhere: a kulhad of hot chai cools faster than a steel tumbler because the earthenware breathes water vapour (extra evaporative cooling); a corpse's core temperature drops from 37 °C toward room temperature at a rate that lets a forensic pathologist estimate time of death; a CPU heatsink is sized so that at full load the chip's temperature equilibrates well below its thermal shutdown threshold.

Pour boiling chai into a kulhad at a Delhi railway station on a January evening. Your phone says 7:10 pm and the tea, just decanted, is very nearly 95 °C — too hot to sip. By 7:13 pm you can start to drink it without flinching; by 7:20 pm it is pleasantly warm; by 7:40 pm, if you forgot to finish it, it is essentially the same temperature as the platform air. Plot the temperature against time and you will notice something specific: the cooling is fast at first (the first 20 °C vanishes in four minutes) and slow at the end (the last 20 °C takes nearly half an hour). The tea does not lose heat at a constant rate. It loses heat at a rate that depends on how much hotter than the air it currently is.

That single observation — cooling rate tracks temperature difference — is Newton's law of cooling. Written as a differential equation it is one of the simplest in physics, and yet it governs a bizarrely wide range of everyday phenomena: the cooling of a forged horseshoe at a Moradabad brassware shop, the slow warming of a bottle of Bisleri left out of the fridge, the steady drop of a dead body's temperature that lets a pathologist read a clock through the skin, and the thermal-throttling behaviour of the laptop on your lap right now. The physics is all the same, and it all falls out of a single first-order differential equation you can solve in three lines.

This article does four things. First, state and justify the law — why the rate of cooling should depend linearly on the temperature difference, and why this is really Stefan's radiation law P \propto T^4 in disguise. Second, solve the differential equation to get the exponential decay T(t) = T_s + (T_0 - T_s)e^{-kt}, step by step, with nothing skipped. Third, show how you would verify the law in a school or JEE-lab setting — the ln-plot trick that turns a curve into a straight line. Fourth, apply the law to the two problems that matter most: how long does hot chai take to become drinkable, and how do forensic pathologists estimate time of death?

Why the rate of cooling should be proportional to \Delta T

A hot object loses heat to its surroundings through three channels: conduction (through whatever solid it touches — a table, your hand, a stand), convection (currents of air or water carry warm fluid away from its surface), and radiation (every surface warmer than 0 K emits electromagnetic radiation, carrying energy). The three channels run in parallel; the total rate of heat loss is the sum.

For each channel, the direction of heat flow is obvious: a hot body loses heat to cooler surroundings. The question is the rate. Does a body that is 50 K hotter than its surroundings lose heat twice as fast as a body 25 K hotter? The answer — approximately yes — is Newton's law.

Here is the qualitative argument for each channel.

Conduction and convection. Through a still air film around the object, heat conducts according to Fourier's law: Q/t = kA \cdot \Delta T / L, which is exactly proportional to \Delta T (with L the film thickness, effectively the boundary-layer depth). Convection — warm air rising off the hot surface — accelerates the transport, but empirically the convective heat-transfer coefficient h_{\text{conv}} for natural convection varies only weakly with \Delta T (typically as \Delta T^{1/4} for a vertical surface in air). For small \Delta T, this weak variation is swamped by the dominant proportionality to \Delta T itself, and the combined conductive–convective loss is well-approximated as h A \Delta T with a single constant h.

Radiation. Every warm surface radiates energy at a rate \sigma \varepsilon A T^4 (Stefan's law), and simultaneously absorbs radiation from its surroundings at a rate \sigma \varepsilon A T_s^4. The net radiative loss is

\frac{Q_{\text{rad}}}{t} = \sigma \varepsilon A \,(T^4 - T_s^4).

This is not linear in \Delta T at first glance. But watch what happens when \Delta T = T - T_s is small. Write T = T_s + \Delta T and expand T^4:

T^4 = (T_s + \Delta T)^4 = T_s^4 + 4 T_s^3 \Delta T + 6 T_s^2 (\Delta T)^2 + \cdots

Why: the binomial expansion (a+b)^n with a = T_s and b = \Delta T. When \Delta T \ll T_s (e.g., 20 K compared to 300 K for room temperature), the quadratic term is 6 \cdot 300^2 \cdot 20^2 \approx 2 \times 10^8, while the linear term is 4 \cdot 300^3 \cdot 20 \approx 2 \times 10^9 — ten times bigger. The linear term dominates.

Keeping only the first-order term:

T^4 - T_s^4 \approx 4 T_s^3 \, \Delta T.

So for small \Delta T, the radiative heat-loss rate becomes

\frac{Q_{\text{rad}}}{t} \approx 4 \sigma \varepsilon A T_s^3 \,\Delta T,

which is linear in \Delta T with an effective coefficient h_{\text{rad}} = 4 \sigma \varepsilon T_s^3. Evaluate at T_s = 300 K: h_{\text{rad}} \approx 4 \cdot (5.67 \times 10^{-8}) \cdot 1 \cdot (300)^3 \approx 6.1 W/(m²·K) — the same order of magnitude as the natural-convection coefficient in air (typically 5–10 W/(m²·K)). At ordinary room temperatures, radiation and natural convection contribute roughly comparable amounts to cooling, and both are linear in \Delta T.

Adding the channels. The total rate of heat loss is the sum:

\frac{Q}{t} = (h_{\text{conv}} + h_{\text{rad}}) \, A \,\Delta T = hA \,\Delta T,

with a combined heat-transfer coefficient h. This is a heat current out of the body, measured in watts.

Connecting to temperature change. The body's temperature drops because its internal energy decreases. If the body has mass m and specific heat c, a loss of heat dQ produces a temperature change dT related by

dQ = -m c \, dT.

Why: Q = mc \Delta T for a temperature change \Delta T at constant specific heat. The minus sign is a sign convention: dQ > 0 here means heat leaving the body, which corresponds to dT < 0 (temperature dropping).

Dividing by dt gives

\frac{dQ}{dt} = -m c \, \frac{dT}{dt}.

Setting the rate at which heat leaves the body (dQ/dt, positive) equal to the surface loss hA\,\Delta T:

-m c \, \frac{dT}{dt} = h A \, (T - T_s).

Rearranging:

\boxed{\;\frac{dT}{dt} = -\frac{hA}{mc}\,(T - T_s) = -k \,(T - T_s),\quad k \equiv \frac{hA}{mc}.\;}

This is Newton's law of cooling in its differential form. The cooling constant k has units of 1/time (seconds⁻¹ or, more conveniently, minutes⁻¹). It is larger for bodies with (i) a large surface-area-to-mass ratio (a thin piece cools faster than a fat one), (ii) small specific heat (metals cool faster than water per unit mass), and (iii) efficient surface heat transfer h (a windy room cools a cup of chai faster than a still one).

A hot body loses heat by conduction (through anything it touches), convection (warm air currents rising off the surface), and radiation (EM emission into the environment). For small temperature differences all three channels are linear in $\Delta T$, so they add to give a total rate proportional to $\Delta T$: the content of Newton's law.

A quick sanity check on the sign

Newton's law is written dT/dt = -k(T - T_s) with a minus sign, and it is worth pausing to see what it says in each case.

If T > T_s (body hotter than surroundings), then T - T_s > 0, and dT/dt < 0: temperature is falling. The body cools, as it should.
If T < T_s (body cooler than surroundings), then T - T_s < 0, and dT/dt > 0: temperature is rising. The body warms up — which is also what happens when a cold bottle of lassi is left on the kitchen counter on a summer afternoon. Newton's law of "cooling" is really a law of approach to thermal equilibrium from either side; the name is historical.
If T = T_s, then dT/dt = 0: the body has equilibrated with its surroundings and no further change occurs.

Solving the differential equation

The equation dT/dt = -k(T - T_s) is a first-order linear ODE with constant coefficients — the simplest differential equation in all of physics, and one of the most useful. You can solve it in three lines by separation of variables.

Step 1. Let \theta(t) = T(t) - T_s. This is the excess temperature over the ambient. Since T_s is constant, d\theta/dt = dT/dt, and the equation becomes

\frac{d\theta}{dt} = -k \theta.

Why: shifting variables from T to \theta = T - T_s removes the constant offset. The equation for \theta is a pure exponential-decay equation — \theta decays toward zero, which corresponds to T decaying toward T_s.

Step 2. Separate variables — put all \theta's on the left and all t's on the right.

\frac{d\theta}{\theta} = -k\, dt.

Why: this is legal because \theta is assumed strictly positive (body hotter than surroundings) during cooling, so dividing by \theta is fine. Physically, you have isolated the rate of fractional change d\theta/\theta and set it equal to -k\,dt, a constant per unit time.

Step 3. Integrate both sides.

\int \frac{d\theta}{\theta} = \int -k \, dt

\ln |\theta| = -k t + C

Why: the antiderivative of 1/\theta is \ln|\theta|, and the antiderivative of -k (a constant) is -kt. The single constant of integration C absorbs both integrals.

Step 4. Exponentiate to isolate \theta.

|\theta| = e^{-kt + C} = e^{C} \cdot e^{-kt}.

Since e^{C} is just another positive constant, rename it A. Dropping the absolute value (we know \theta > 0 during cooling):

\theta(t) = A \, e^{-kt}.

Step 5. Apply the initial condition. At t = 0, the body's temperature is T_0, so \theta(0) = T_0 - T_s. Substituting:

\theta(0) = A \cdot e^{0} = A = T_0 - T_s.

Why: e^{0} = 1, so the constant A is just the initial excess temperature. This is the standard exponential-decay setup: at t=0, the "amount" is whatever you started with; at later times, it is the starting amount multiplied by e^{-kt}.

Step 6. Translate back to T. Since \theta = T - T_s,

T(t) - T_s = (T_0 - T_s) \, e^{-kt},

\boxed{\;T(t) = T_s + (T_0 - T_s) \, e^{-kt}.\;}

Three features of this solution are worth grooving into your intuition.

At t = 0, T = T_s + (T_0 - T_s) = T_0. Correct initial condition.
As t \to \infty, e^{-kt} \to 0, and T \to T_s. The body equilibrates with its surroundings at long times. However long you wait, the body never quite reaches T_s — it only gets arbitrarily close, because the exponential never vanishes. In practice, after about 5/k seconds the body is at T_s to within less than 1% of the original excess.
The time constant \tau = 1/k is the characteristic cooling time. After one \tau, the excess temperature has fallen to 1/e \approx 36.8\% of its initial value. After two \tau, it is at 1/e^2 \approx 13.5\%. After five \tau, it is less than 1%.

The time constant in numbers

For a cup of tea in air at 25 °C, typical values are m = 0.2 kg, c = 4186 J/(kg·K) (water), A \approx 0.015 m² (surface area of the top plus sides of the mug above the water), and h \approx 15 W/(m²·K) (natural convection plus radiation from a warm surface). Then

k = \frac{hA}{mc} = \frac{15 \cdot 0.015}{0.2 \cdot 4186} = \frac{0.225}{837.2} \approx 2.7 \times 10^{-4}\ \text{s}^{-1} \approx 0.016\ \text{min}^{-1}.

So \tau = 1/k \approx 3700 s \approx 62 min. Starting at 95 °C in a 25 °C room, after 62 minutes the excess is down to 70 \cdot 0.368 \approx 26 °C, so T \approx 51 °C. That matches ordinary experience — a cup of tea left on a table is undrinkably cold after about an hour.

Notice the ingredients: the time constant is long because water has a huge specific heat. The same steel of tea (pardon the phrasing) held in a hypothetical liquid with c = 1000 J/(kg·K) would cool four times faster. Hence the cultural wisdom: a kulhad of chai, full of water, takes its time.

Visualising the decay

The exponential decay $T(t) = T_s + (T_0 - T_s)\,e^{-kt}$. The curve falls steeply at first (large slope, proportional to the initial excess $T_0 - T_s$), then flattens as $T$ approaches $T_s$. At $t = \tau = 1/k$, the excess has fallen to $36.8\%$ of its initial value. The ambient $T_s$ is an asymptote — reached in the limit, never at any finite time.

The instantaneous slope of the curve is always -k\Delta T

Notice something beautiful in the picture above. At any point on the curve, the slope dT/dt equals -k times the vertical distance from that point down to the ambient asymptote. The steeper the distance, the steeper the slope; the closer you get to the asymptote, the gentler the slope. This is exactly what the differential equation says — and if you mentally connect the slopes along the curve, you get a tangent-line family whose envelope is the exponential.

This is why cooling slows down so dramatically as the body approaches equilibrium. The first 20 °C of excess temperature disappears while the body is hot and losing energy vigorously; the last 1 °C of excess takes nearly as long again, because at that point the rate of heat loss is 95% less than it was at the start.

Explore the cooling curve

Drag the red point along the horizontal axis to change the cooling constant $k$ from 0.02 to 0.30 min⁻¹. Watch how the curve responds: a larger $k$ means a steeper decay and a shorter time constant $\tau$. The temperature at 30 minutes is shown in the readout — at $k = 0.08$ min⁻¹ (a small mug of chai), tea starting at 95 °C in a 25 °C room is at about 31 °C after half an hour.

Experimental verification — the ln-plot trick

How would you actually check Newton's law in a lab? Heat water to nearly boiling, pour it into a beaker, drop in a thermometer, and record T every 30 seconds as it cools. You will get a smooth decaying curve. But is it truly exponential, or just curvy and vaguely decaying? The eye is a bad judge of exponential-versus-something-else when looking at a single curve.

The trick is to linearise. If Newton's law holds,

T - T_s = (T_0 - T_s) \, e^{-kt}.

Take the natural logarithm of both sides:

\ln(T - T_s) = \ln(T_0 - T_s) - k t.

Why: \ln(ab) = \ln a + \ln b, and \ln(e^{-kt}) = -kt. The constant \ln(T_0 - T_s) is just a number — the intercept of the line.

Now plot \ln(T - T_s) on the vertical axis against t on the horizontal. If Newton's law is right, you get a straight line of slope -k and intercept \ln(T_0 - T_s). If the points deviate systematically from a line, the law is failing — typically because the assumption of small \Delta T or constant ambient has been violated.

A sample dataset from a school-lab experiment:

t (min)	T (°C)	T - T_s (°C, T_s = 25)	\ln(T - T_s)
0	95	70	4.248
2	83	58	4.060
4	73	48	3.871
6	65	40	3.689
8	58	33	3.497
10	53	28	3.332
15	42	17	2.833
20	35	10	2.303

Plotting column 4 against column 1 gives a line of slope \approx -0.098 min⁻¹ — so k \approx 0.10 min⁻¹ and the time constant \tau \approx 10 min. The straightness of the line is the verification. If you are doing this in a JEE lab, the data will never be perfectly linear at both ends — sometimes the first few points curve because the thermometer has not yet equilibrated with the water (instrumental lag), and sometimes the last few points curve because T is close to T_s and small measurement errors in T - T_s blow up under the logarithm. The middle region is usually clean.

Left: raw data $T(t)$ looks exponential-ish. Right: plotting $\ln(T - T_s)$ against $t$ straightens the curve. The slope of this straight line is $-k$; its intercept is $\ln(T_0 - T_s)$. A straight-line fit is how Newton's law is verified experimentally.

Worked examples

Example 1: A cup of chai on the station platform

A vendor at Old Delhi station pours freshly-boiled chai into your steel tumbler at 90 °C. The platform air is at 15 °C on a cool December evening. Five minutes later, the chai has cooled to 70 °C. How long (from the moment it was poured) will it take to cool to a comfortable drinking temperature of 50 °C?

Three snapshots of the same tumbler on the same platform. From the first two you extract $k$; the third is what you're solving for.

Step 1. Identify the knowns. Initial temperature T_0 = 90 °C. Ambient T_s = 15 °C. At t_1 = 5 min, T_1 = 70 °C. Target: find t_2 when T_2 = 50 °C.

Why: with three data points (initial, known time, target time) and one unknown constant k, the system is completely determined. Extract k from the known-time data, then use it to predict the target time.

Step 2. Plug into Newton's solution at t = 5 min to extract k.

T_1 = T_s + (T_0 - T_s)\, e^{-k t_1}

70 = 15 + (90 - 15)\, e^{-5k}

55 = 75 \, e^{-5k}

e^{-5k} = \frac{55}{75} = \frac{11}{15} \approx 0.7333

Why: isolating the exponential by subtracting T_s = 15 from both sides and dividing by the initial excess T_0 - T_s = 75 gives the fractional excess remaining — 11/15 after 5 minutes.

Step 3. Take the natural log of both sides.

-5k = \ln(11/15) \approx \ln(0.7333) \approx -0.3102

k \approx \frac{0.3102}{5} \approx 0.0620 \text{ min}^{-1}

Why: \ln is the inverse of the exponential. Dividing by 5 gives k with units of (minutes)⁻¹ because that is the time unit I used.

Step 4. Now solve for t_2 when T_2 = 50 °C.

50 = 15 + 75 \, e^{-k t_2}

35 = 75 \, e^{-k t_2}

e^{-k t_2} = \frac{35}{75} = \frac{7}{15} \approx 0.4667

-k t_2 = \ln(0.4667) \approx -0.7621

t_2 = \frac{0.7621}{0.0620} \approx 12.3 \text{ min}

Why: same procedure as Step 2 but in reverse — you know the final T and want to find the time it takes. The excess to dissipate is 35/75 = 7/15 of the initial excess, which requires \ln(15/7) / k minutes.

Result: The chai takes about 12.3 minutes (from the moment it was poured) to cool from 90 °C to 50 °C. Since 5 of those minutes have already passed, you have about 7 more minutes to wait.

What this shows: Once you extract k from a single pair of measurements, the entire cooling curve is determined. Newton's law is a one-parameter model — measure the cooling rate at any one excess temperature and the whole history is predicted.

Example 2: The forensic time-of-death problem

A body is discovered in a Mumbai apartment at 7:00 am, and its core temperature is measured at 30 °C. One hour later, at 8:00 am, the core temperature has dropped to 29 °C. The room has been kept at a steady 22 °C by air-conditioning. Assuming a normal core body temperature of 37 °C at the moment of death, and that Newton's law of cooling applied from then on, estimate the time of death.

Core body temperature decays from 37 °C at the moment of death toward the ambient 22 °C. Two measurements — 30 °C at 7 a.m. and 29 °C at 8 a.m. — fix both $k$ and the time of death.

Step 1. Set up coordinates. Let t = 0 correspond to 7:00 am. Then 8:00 am is t = 1 hour, and the (unknown) moment of death is t = -\Delta hours before 7 am, where \Delta > 0 is what we are solving for.

At t = -\Delta: T = 37 °C. At t = 0: T = 30 °C. At t = 1 h: T = 29 °C. Ambient T_s = 22 °C throughout.

Why: aligning t = 0 with a known measurement makes the algebra cleaner. The death time becomes a single unknown \Delta that shifts the whole cooling curve.

Step 2. Find k using the two measurements after discovery.

T(0) - T_s = (T_0 - T_s) \, e^{-k \cdot 0} \text{ — but we don't know } T_0 \text{ (the temp at death).}

So I use the ratio between 7 am and 8 am instead:

\frac{T(1) - T_s}{T(0) - T_s} = \frac{(T_0 - T_s) e^{-k \cdot 1}}{(T_0 - T_s) e^{-k \cdot 0}} = e^{-k}.

\frac{29 - 22}{30 - 22} = e^{-k}

\frac{7}{8} = e^{-k}

k = -\ln(7/8) = \ln(8/7) \approx 0.1335 \text{ h}^{-1}.

Why: taking the ratio of two measurements of T - T_s at different times eliminates the unknown initial excess T_0 - T_s entirely. What remains is the pure exponential factor e^{-k \Delta t}, from which k drops out directly.

Step 3. Use the cooling equation from the moment of death (t = -\Delta, T = 37 °C) to the first measurement (t = 0, T = 30 °C).

T(0) - T_s = (37 - T_s) \, e^{-k \cdot \Delta}

30 - 22 = (37 - 22) \, e^{-k \Delta}

8 = 15 \, e^{-k \Delta}

e^{-k \Delta} = \frac{8}{15}

-k \Delta = \ln(8/15) \approx -0.6286

\Delta = \frac{0.6286}{0.1335} \approx 4.71 \text{ h}

Why: the time elapsed from death to the first measurement is whatever makes the excess temperature decay from (37 - 22) = 15 °C down to (30 - 22) = 8 °C. That is \ln(15/8)/k.

Step 4. Convert \Delta hours back into clock time.

7:00 am minus 4.71 hours = 7:00 am minus 4 hours 43 minutes \approx 2:17 am.

Result: The estimated time of death is approximately 2:17 am.

What this shows: Newton's law gives forensic pathologists a clock they can read by measuring a single temperature now and a follow-up an hour later. In practice, the estimate is corrected for clothing, body mass, posture, and ambient humidity, but the core physics — exponential approach to ambient — is exactly what is used here. Real-world uncertainty is typically \pm 2 hours; this is why TV shows never quote times of death to the minute.

Common confusions

"The rate of cooling is constant." Absolutely not — and believing this leads to predictions that are badly wrong. A cup of tea starting at 95 °C in a 25 °C room loses heat at, say, 15 W initially (while the excess is 70 K); after the tea has cooled to 40 °C, the excess is 15 K, and the rate has fallen to about 15 · (15/70) \approx 3.2 W. The rate of cooling slows down as the body approaches equilibrium. A constant-rate-of-cooling model is a straight line; Newton's law is a curve that flattens.
"After time \tau = 1/k, the body has reached ambient temperature." No. After \tau, the excess has decayed to 1/e \approx 37\% of its initial value, not zero. The body is still warmer than the surroundings by 37% of the original excess. Full equilibration (within 1%) takes about 5\tau. Exponentials never quite reach their asymptote.
"Newton's law and Stefan's law contradict each other." They don't. Stefan's law P \propto T^4 is the exact radiation physics. Newton's law is the linearisation of Stefan's law (plus convection and conduction) valid when \Delta T \ll T_s. For red-hot steel at 1000 K in a 300 K room, \Delta T = 700 K is not small compared to T_s, and the linear approximation fails — Stefan's law predicts a much faster cooling than Newton's. For warm tea at 350 K in a 300 K room, \Delta T = 50 K is small, and Newton's law is accurate to a few percent.
"The cooling constant k is a universal property of the material." No — it lumps together the body's size, shape, surface, specific heat, and the environment's convective and radiative heat-transfer properties. The same mug of tea in still air versus in front of a fan will have very different k values. Tabulated values of k are specific to a particular apparatus and environment.
"A small object at the same temperature as a large one cools at the same rate." No — small objects cool faster. Look at k = hA/(mc): for a sphere, mass scales as R^3 but surface area as R^2, so A/m \propto 1/R. A small ice cube disappears fast; a glacier takes millennia. This is why ISRO's tiny nanosatellites need precise thermal design — they have huge surface-to-mass ratios and lose or gain heat very quickly.

If you came here to know what Newton's law says, use it to solve problems, and leave — you have what you need. What follows is for readers who want the full connection to Stefan's law, the multi-channel derivation of the effective k, and the non-isothermal generalisation used in real thermal engineering.

Full derivation from Stefan's law plus convection

The radiative heat loss from a surface at temperature T into surroundings at T_s is

P_{\text{rad}} = \sigma \varepsilon A \, (T^4 - T_s^4),

with \sigma = 5.67 \times 10^{-8} W/(m²·K⁴) (Stefan–Boltzmann constant) and \varepsilon the surface emissivity (0 < \varepsilon \le 1). Factor out T_s^4:

P_{\text{rad}} = \sigma \varepsilon A T_s^4 \left[\left(\frac{T}{T_s}\right)^4 - 1\right].

Let u = (T - T_s)/T_s, so T/T_s = 1 + u, and expand:

\left(1 + u\right)^4 - 1 = 4u + 6u^2 + 4u^3 + u^4.

For |u| \ll 1, keep only the linear term:

P_{\text{rad}} \approx 4 \sigma \varepsilon A T_s^3 \, (T - T_s) = h_{\text{rad}} A (T - T_s),

with

h_{\text{rad}} = 4 \sigma \varepsilon T_s^3.

This is the effective linear "radiative coefficient" valid near T = T_s. At T_s = 300 K and \varepsilon = 0.9 (a matte painted surface), h_{\text{rad}} \approx 4 \cdot 5.67 \times 10^{-8} \cdot 0.9 \cdot 2.7 \times 10^{7} \approx 5.5 W/(m²·K).

Convective heat loss, for natural convection from a warm vertical surface in air, is empirically

P_{\text{conv}} = h_{\text{conv}} A (T - T_s),

with h_{\text{conv}} typically in the 2–25 W/(m²·K) range depending on geometry. For moderately-warm surfaces at room temperature, h_{\text{conv}} \approx 5–10 W/(m²·K), the same order as h_{\text{rad}}. Adding:

P_{\text{total}} = (h_{\text{rad}} + h_{\text{conv}}) A (T - T_s) = h_{\text{eff}} A (T - T_s).

Dividing by -mc (to convert heat loss into temperature drop) yields Newton's equation

\frac{dT}{dt} = -\frac{h_{\text{eff}} A}{mc}(T - T_s) = -k (T - T_s).

The effective k is thus built from four ingredients: radiative coefficient (\varepsilon T_s^3), convective coefficient (h_{\text{conv}}), geometric ratio (A/m), and thermal mass (c).

Range of validity

Compare the linear and quadratic terms in the expansion of (1+u)^4. The linear term is 4u; the quadratic is 6u^2. The quadratic is less than 10% of the linear when 6u^2 < 0.4 u, i.e. u < 0.067. For T_s = 300 K, that means T - T_s < 20 K. For \Delta T up to 20–30 K, Newton's law is accurate to a few percent; for \Delta T = 100 K, the error is already tens of percent; for red-hot iron at 800 K in a 300 K room, Newton's law is useless.

Time-varying ambient: the generalisation

The derivation above assumed T_s is constant. What if the surroundings themselves are warming or cooling — for instance, a baby's bathwater cooling in a bathroom whose temperature is rising as the hot shower in the next room heats the walls? The equation becomes

\frac{dT}{dt} + k T = k T_s(t),

which is a first-order linear ODE with a time-dependent forcing term. The solution is the convolution

T(t) = T_0 e^{-kt} + k \int_0^t e^{-k(t-t')} T_s(t') \, dt'.

Why: this is the standard integrating-factor solution for a linear first-order ODE. The first term is the homogeneous decay from the initial condition; the second term is the response to the forcing, weighted by an exponential memory kernel that forgets the distant past.

In the limit where T_s(t) changes slowly compared to 1/k (the body can keep up), T(t) \approx T_s(t): the body tracks the ambient. In the opposite limit where T_s oscillates rapidly, T barely responds: the body averages out the fast fluctuations. This is why a thick-walled thermos keeps coffee hot on a moving train (fast thermal fluctuations of the cabin average to zero) but fails on a day-long journey (slow drifts of ambient eventually win).

The lumped-capacity assumption

Newton's law assumes the body's temperature is uniform — a single number T that evolves in time. This is the "lumped capacity" or "well-mixed" assumption. It fails when the body is a poor internal conductor and takes longer to equilibrate internally than it does to exchange heat with the surroundings. The dimensionless parameter that decides is the Biot number:

\text{Bi} = \frac{h L_c}{k_{\text{material}}},

where L_c is a characteristic internal length (volume over surface area for a general body) and k_{\text{material}} is the material's thermal conductivity. When \text{Bi} \ll 1 (small body, good conductor, gentle surface exchange), the lumped-capacity model works and Newton's law applies. When \text{Bi} \gg 1 (large body, poor conductor, aggressive surface exchange), internal temperature gradients matter and you need the full heat equation \partial T / \partial t = \alpha \nabla^2 T instead. A steel bar in a still room has \text{Bi} \sim 0.01 — Newton's law works perfectly. A brick wall in a wind has \text{Bi} \sim 10 — you need the full heat equation.

The thermal time constant and circuit analogy

Newton's law has an exact electrical analogue. The "thermal capacitor" is the body with thermal capacitance C_{\text{th}} = mc (joules per kelvin), charged with a quantity of heat Q = C_{\text{th}} (T - T_s). The "thermal resistor" between the body and the ambient is R_{\text{th}} = 1/(hA) (kelvins per watt). The product R_{\text{th}} C_{\text{th}} = mc/(hA) = 1/k = \tau is the thermal time constant — identical in form to the electrical RC time constant. A cooling body is just a charged thermal capacitor discharging through a thermal resistor.

This analogy is the designer's tool. A CPU heatsink is chosen so that \tau_{\text{CPU}} = R_{\text{sink,th}} \cdot C_{\text{CPU}} is short enough that a burst of power dissipation doesn't drive the chip temperature past its thermal limit before the heat can flow out. A wine cellar's walls are sized so that \tau_{\text{cellar}} is long — many hours or days — so that fast daily outside-temperature swings barely perturb the cellar's temperature. Same equation, opposite design goals.

Why a kulhad cools chai faster than a steel tumbler

Two vessels of the same size and shape, filled with the same hot chai, at the same ambient. Why does the unglazed kulhad of earthenware cool tea noticeably faster than a polished steel tumbler? The pure Newton's-law constants favour the steel tumbler as a slower cooler (higher k_{\text{material}} means heat conducts out the walls faster), but the kulhad wins. The reason is evaporation. Unglazed earthenware is porous; water from inside the chai seeps slowly into the clay and evaporates from the outer surface. Every gram of water that evaporates carries off about 2260 J of latent heat — an enormous extra cooling channel not included in the Newton's-law model. The effective k of the kulhad includes this evaporative contribution, and can easily be double the pure-radiative-plus-convective value of the steel tumbler. This is why the railway-platform kulhad works: it uses the same physics as a ghada (the traditional clay water pot that keeps drinking water cool in summer) — evaporation through the pores is a refrigerator for the vessel's contents.

Connection to the heat equation

Newton's law is a particular solution of the full partial differential equation of heat conduction — the one-dimensional heat equation:

\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2},

with appropriate boundary conditions. For a uniformly-heated body with a convective boundary condition -k_{\text{material}} \partial T/\partial x = h (T_{\text{surf}} - T_s), in the low-Biot limit where the body is internally uniform, the heat equation integrates to Newton's law. The full solution involves an infinite series of decaying eigenmodes; the lumped-capacity limit keeps only the slowest mode (the pure exponential), and that one mode is Newton's law.

Non-linear corrections for high \Delta T

For larger temperature differences, one can retain more terms of the Stefan-law expansion. To next order:

P_{\text{rad}} \approx 4 \sigma \varepsilon A T_s^3 \, \Delta T \left(1 + \frac{3 \Delta T}{2 T_s} + \cdots \right).

The bracketed correction is the ratio of the quadratic and linear contributions. At \Delta T = T_s (body twice as hot as ambient in absolute terms), the correction factor is 1 + 3/2 = 2.5 — Newton's law underestimates cooling by a factor of 2.5. The full Stefan's law must be used for hot iron forgings, glass-blowing, or the early stages of CPU cooling before the chip has approached room temperature.

Where this leads next

Radiation — Stefan's T^4 law, the full physics from which Newton's law is a linearisation.
Convection — how warm fluid motion carries heat away from a surface, supplying one of the two components of the effective k.
Conduction — Fourier's law, heat flow through still media, the microscopic picture behind the convective boundary layer.
Kinetic Theory — Pressure and Temperature — the molecular origin of temperature itself, which makes "temperature difference" a microscopically meaningful quantity.
Exponential Function — the mathematics of e^{-kt}, the universal decay curve that governs cooling, radioactive decay, RC circuits, and chemical reactions.