Newton's Law of Cooling

Forensic science estimates time of death this way. A body starts near 37 °C and cools towards the room. If the room is at 20 °C and the corpse is found at 30 °C, then with a cooling constant $k = 0.10\ \mathrm{h^{-1}}$ measured for the conditions, equation (3a) inverts to $t = \ln\!\big((37-20)/(30-20)\big)/k = \ln(1.7)/0.10 = 5.3$ hours since death. Pathologists correct for clothing, body mass and airflow — all adjustments to $h$, $A$, $m$ and $c$ inside $k = hA/(mc)$.

Electronics and engines are designed around the same equation: a CPU heat sink maximises $A$ and, with its fan, $h$, pushing $k$ up so the chip settles at a safe temperature. Food-safety rules are cooling rules — hot food must cross the bacterial danger zone fast, so kitchens divide a stockpot into shallow trays, multiplying $A/m$ and therefore $k$. And the school cooling-curve experiment remains one of the cleanest demonstrations that a real system obeys an exponential law.

Put a hot metal block in a cool room. Heat leaks from its surface by three routes at once: conduction into whatever it touches, convection as the neighbouring air warms, thins and floats away, and thermal radiation streaming off as infrared. Every one of those routes is driven by the difference between block and surroundings. Nothing here depends on the block's temperature in isolation: a block at 300 °C in a 290 °C oven barely cools at all, while a block at 30 °C on a −10 °C night bleeds heat steadily. What matters is the excess, $T - T_s$. Now watch that unfold in time. At the start the excess is large, so heat pours out and the temperature plunges. But every degree lost shrinks the excess, a smaller excess means a slower leak, and so the next degree takes longer to lose. The body throttles its own cooling. That feedback — the driver of a process being consumed by the process itself — is the standard recipe for exponential decay, the same shape as a discharging capacitor or a decaying nucleus. The curve dives, bends over, and settles onto the line $T = T_s$ like a plane easing onto a runway.

Read that figure once and the topic is yours. The first ten minutes cost the coffee 25 °C, the next ten only 15.4 °C, the next only 9.5 °C — and each drop is 0.615 times the one before, the same ratio every time. A constant ratio per fixed interval is the fingerprint of an exponential, and it is what separates the student who understands cooling from the one who assumes (wrongly, and expensively) that a body losing 25 °C in the first ten minutes must lose another 25 °C in the next.

Newton's statement, for a body at temperature $T$ in surroundings held at $T_s$:

where: $T$ — the body's temperature at some instant; $T_s$ — the constant temperature of the surroundings (some textbooks write $\theta_0$); $T - T_s$ — the temperature excess, the only quantity the law cares about; $k$ — the cooling constant, in $\mathrm{min^{-1}}$ or $\mathrm{s^{-1}}$, a property of this body in this environment, not of the material alone; $dT/dt$ — the rate at which the temperature changes. The minus sign is bookkeeping: a hotter body has a positive excess, so $dT/dt$ comes out negative — the temperature is falling. You do not need calculus to read equation (1). In words: the number of degrees lost per minute is proportional to how many degrees the body stands above the room. Pictorially, $dT/dt$ is the slope of the tangent to the cooling curve — steep at the start, shallow at the end — and the law says that slope is always the same fixed multiple of the vertical gap between curve and ambient line. For working problems, boards use the finite-difference form. Take a small drop from $T_1$ to $T_2$ in a time $\Delta t$. The average cooling rate over that window is $(T_1 - T_2)/\Delta t$, and the excess to pair with it is that of the mean temperature of the interval:

where: $T_1$, $T_2$ — the temperatures at the start and end of the interval; $\Delta t$ — the time taken; $(T_1+T_2)/2$ — the mean temperature, standing in for the body's steadily changing $T$; $k$ — the same cooling constant. The physical reading: across a small window the body is always somewhere between $T_1$ and $T_2$, so the honest number to represent it is the middle one. Equation (2) is the average-temperature method — pure algebra, no calculus — and CBSE, ICSE, NEET and JEE numericals are built on it. Its one condition: the drop must be small compared with the excess.

Feed equation (1) into the machinery of calculus and out comes a formula for the temperature at any time. That solving step belongs to a later course; the answer belongs to you now, because boards quote it and competitive papers expect you to wield it:

where: $T_0$ — the body's temperature at $t = 0$; $T_0 - T_s$ — the initial excess; $e \approx 2.718$; $k$ — the cooling constant of equation (1); $T$ — the temperature after a time $t$. Read the structure, not the symbols: the excess starts at $T_0 - T_s$ and is multiplied by the shrinking factor $e^{-kt}$. Rearranged, that is the sentence worth memorising:

The excess decays exponentially; everything else is decoration. Two consequences follow, and both are exam questions in disguise.

Equal times give equal fractions. In any fixed interval $\Delta t$ the excess is multiplied by $e^{-k\Delta t}$ — the same factor wherever you start. Our coffee's excess is multiplied by $e^{-0.0486 \times 10} = 0.615$ every ten minutes. From an initial excess of $65\ \mathrm{^\circ C}$: after 10 min it is $65 \times 0.615 = 40\ \mathrm{^\circ C}$ (temperature 65 °C), after 20 min $65 \times 0.615^2 = 24.6\ \mathrm{^\circ C}$ (49.6 °C), after 30 min $65 \times 0.615^3 = 15.1\ \mathrm{^\circ C}$ (40.1 °C). The absolute drops shrink — 25.0, then 15.4, then 9.5 °C — while the fraction stays locked at 0.615. Any "cools from A to B in $t$ minutes; find the temperature after another $t$ minutes" question tests this and nothing else. It also means the excess has a half-life, $t_{1/2} = \ln 2/k = 14.3$ min, and a time constant $\tau = 1/k = 20.6$ min in which it falls to $1/e = 36.8\%$ of its value.

The body never reaches ambient, and never dips below it. As $t$ grows, $e^{-kt}$ shrinks towards zero but is never zero, so $T$ creeps onto $T_s$ and stops. Cooling cannot overshoot: at $T = T_s$ the excess is zero, so by equation (1) the rate is zero. That is thermal equilibrium, and it is why the curve flattens instead of crossing.

The school experiment is simple: fill a calorimeter with hot water and record its temperature each minute as it cools. Plotting $T$ against $t$ gives the falling curve of the first figure — pretty, but not proof, since many curves fall like that. The proof comes from replotting. Take successive pairs of readings, compute the cooling rate $(T_1 - T_2)/\Delta t$ for each, and plot it against the excess of the mean temperature, $\overline{T} - T_s$. Newton's law predicts equation (2): a straight line through the origin, of slope $k$. That is the graph that earns the marks.

Three marks hide in that graph. The line must pass through the origin, because zero excess means zero cooling — a best-fit line with an intercept means your $T_s$ is wrong. The slope is $k$: a rate of $3.0\ \mathrm{^\circ C/min}$ at an excess of $60\ \mathrm{^\circ C}$ gives $k = 3.0/60 = 0.05\ \mathrm{min^{-1}}$, so at an excess of $24\ \mathrm{^\circ C}$ the same body must cool at $0.05 \times 24 = 1.2\ \mathrm{^\circ C/min}$. And the real points bend above the line at large excess (the red dashed curve), because a body far above ambient loses a disproportionate share by radiation, which follows $T^4$ rather than the excess — precisely what Dulong and Petit found in 1817.

Newton's law is not fundamental but an approximation, and knowing which approximation is what lifts an answer into the top band. A hot body loses heat two main ways. Convection genuinely is close to linear in the excess: the surface sheds power $H = hA(T - T_s)$, where $h$ is the convective heat-transfer coefficient ($5$–$25\ \mathrm{W/m^2\,K}$ in still air, far more with a fan). That escaping power drains a store of heat capacity $mc$, so the temperature falls at $H/(mc)$ — which identifies the cooling constant itself:

where: $h$ — the surface coefficient, set by airflow and surface; $A$ — the exposed area; $m$ — the mass; $c$ — the specific heat capacity. Read it as a contest between losing (the numerator, all surface) and storing (the denominator, all bulk): a thin sheet or a droplet has a large $k$ and cools in moments, a bathtub has a tiny $k$ and stays warm for hours. It also settles every "which body cools faster" MCQ, and it shows why $k$ is a property of the whole set-up rather than of the material.

Radiation, the second route, is emphatically not linear. The Stefan–Boltzmann law gives the net radiated power as

where: $e$ — emissivity (0 to 1); $\sigma = 5.67\times10^{-8}\ \mathrm{W/m^2\,K^4}$ — the Stefan–Boltzmann constant; $A$ — surface area; $T$, $T_s$ — the absolute temperatures of body and surroundings, in kelvin. That fourth power is ferocious and looks nothing like a straight line in the excess. Yet for a small excess it flattens into one — and that is the deep reason Newton's law exists at all.

Board level — the next ten minutes of cooling (CBSE/ICSE/NEET pattern)

A body cools from $60\ \mathrm{^\circ C}$ to $50\ \mathrm{^\circ C}$ in 10 minutes. The surroundings are at $25\ \mathrm{^\circ C}$. What is its temperature after the next 10 minutes?

Step 1. First interval, by equation (2): the drop is $10\ \mathrm{^\circ C}$ in 10 min, so the rate is $10/10 = 1.0\ \mathrm{^\circ C/min}$; the mean temperature is $(60+50)/2 = 55\ \mathrm{^\circ C}$, so the excess is $55 - 25 = 30\ \mathrm{^\circ C}$.

Step 2. Hence $1.0 = k \times 30$, giving $k = 1/30 = 0.0333\ \mathrm{min^{-1}}$.

Step 3. Second interval, with the unknown final temperature $T$: the rate is $(50 - T)/10$ and the mean temperature is $(50+T)/2$, so equation (2) reads $\dfrac{50 - T}{10} = \dfrac{1}{30}\left(\dfrac{50+T}{2} - 25\right)$.

Step 4. Multiply through by 30: $3(50 - T) = \dfrac{50+T}{2} - 25$, and the $25$ cancels the $50/2$ on the right, leaving $150 - 3T = \dfrac{T}{2}$.

Step 5. Solve: $150 = \dfrac{7T}{2}$, so $T = \dfrac{300}{7} = \boxed{42.9\ \mathrm{^\circ C}}$ (quoted in many keys as $42.85\ \mathrm{^\circ C}$). Sense-check: the second drop is $50 - 42.9 = 7.1\ \mathrm{^\circ C}$, smaller than the first drop of $10\ \mathrm{^\circ C}$, exactly as a shrinking excess demands. The candidate who assumes a second $10\ \mathrm{^\circ C}$ drop and writes $40\ \mathrm{^\circ C}$ loses the whole mark — and that is the commonest wrong answer in the topic.

Competitive level — coffee, an exponential and a clock (JEE-Main/JEE-Advanced pattern)

Coffee at $90\ \mathrm{^\circ C}$ stands in a room at $25\ \mathrm{^\circ C}$ and cools to $65\ \mathrm{^\circ C}$ in 10 minutes. Find (a) the cooling constant, (b) the temperature 20 minutes after the start, and (c) when the coffee reaches a drinkable $40\ \mathrm{^\circ C}$.

Step 1. Work in excesses, never raw temperatures — that is the whole trick. Initial excess $= 90 - 25 = 65\ \mathrm{^\circ C}$; excess after 10 min $= 65 - 25 = 40\ \mathrm{^\circ C}$.

Step 2. Equation (3a) gives $40 = 65\,e^{-10k}$, so $e^{-10k} = 40/65 = 0.6154$ and $k = \dfrac{\ln(65/40)}{10} = \boxed{0.0486\ \mathrm{min^{-1}}}$. Writing $k$ as $\ln(65/40)/10$ rather than $-\ln(40/65)/10$ saves you from a stray minus sign.

Step 3. Part (b) needs no exponential at all. Every 10-minute window multiplies the excess by the same $0.6154$, so after 20 min the excess is $65 \times 0.6154^2 = 24.6\ \mathrm{^\circ C}$ and the temperature is $T = 25 + 24.6 = \boxed{49.6\ \mathrm{^\circ C}}$. The long route, $T = 25 + 65\,e^{-0.0486 \times 20}$, returns the identical $49.6\ \mathrm{^\circ C}$ — but the fractional route needs no calculator.

Step 4. Part (c): invert equation (3a). The target excess is $40 - 25 = 15\ \mathrm{^\circ C}$, so $15 = 65\,e^{-kt}$ and $t = \dfrac{\ln(65/15)}{k} = \dfrac{1.4663}{0.04855} = \boxed{30.2\ \text{minutes}}$. Timing note: about two minutes for all three parts. The exam-winning move is Step 3 — spotting that "another 10 minutes" means "multiply the excess by the same factor again" converts a logarithm into a single multiplication. The classic wasted minute is converting to kelvin: don't. Every temperature here sits inside a difference, and a difference is numerically identical on both scales.

Competitive level — find the temperature of the room (JEE-Advanced/NEET pattern)

A body cools from $80\ \mathrm{^\circ C}$ to $64\ \mathrm{^\circ C}$ in 5 minutes, and from $64\ \mathrm{^\circ C}$ to $52\ \mathrm{^\circ C}$ in the next 5 minutes. Find the temperature of the surroundings.

Step 1. Recognise the type: two intervals, equal times, unequal drops, and the unknown is $T_s$ itself — so $T_s$ must remain a symbol, $\theta$, inside every excess.

Step 2. First interval: drop $16\ \mathrm{^\circ C}$ in 5 min, so rate$_1 = 3.2\ \mathrm{^\circ C/min}$; mean temperature $(80+64)/2 = 72\ \mathrm{^\circ C}$; excess $72 - \theta$.

Step 3. Second interval: drop $12\ \mathrm{^\circ C}$ in 5 min, so rate$_2 = 2.4\ \mathrm{^\circ C/min}$; mean temperature $(64+52)/2 = 58\ \mathrm{^\circ C}$; excess $58 - \theta$.

Step 4. Divide equation (2) for the first interval by equation (2) for the second, and $k$ cancels: $\dfrac{3.2}{2.4} = \dfrac{72 - \theta}{58 - \theta}$, i.e. $\dfrac{4}{3} = \dfrac{72 - \theta}{58 - \theta}$. Cross-multiplying, $4(58 - \theta) = 3(72 - \theta)$, so $232 - 4\theta = 216 - 3\theta$ and $\boxed{T_s = 16\ \mathrm{^\circ C}}$.

Step 5. Cross-check with the exponential — this problem is a beautiful special case. With $T_s = 16$ the three excesses are $64$, $48$ and $36\ \mathrm{^\circ C}$, and $48/64 = 0.75$ while $36/48 = 0.75$ too. Equal fractional drops in equal times: the exponential signature, satisfied exactly, so both methods agree perfectly on $16\ \mathrm{^\circ C}$. They differ only on $k$ — the ratio method gives $3.2/56 = 0.0571\ \mathrm{min^{-1}}$, the exponential $\ln(4/3)/5 = 0.0575\ \mathrm{min^{-1}}$, a $0.7\%$ gap invisible in any exam. Timing note: ninety seconds. Everything hinges on Step 4 — divide the two equations so $k$ vanishes. A student who instead writes two exponentials and takes logs burns four minutes to reach the same $16\ \mathrm{^\circ C}$. Note too what makes the problem solvable: the drops in equal times are unequal, and their ratio is what pins down $T_s$.

Using the temperature instead of the excess. Every bracket in this topic is $(T - T_s)$. NEET assertion-reason questions bait exactly this: "a body at 60 °C always cools at the same rate" is false, because the rate depends on the room it sits in. If $T_s$ is given, it appears in your working.

Assuming equal drops in equal times. Cooling is exponential: equal intervals give equal fractions, not equal amounts. A body falling 60 → 50 °C in ten minutes does not reach 40 °C in the next ten — it reaches 42.9 °C. This is the most-marked error in CBSE and ICSE papers.

Using the starting temperature in the average-temperature method. Equation (2) needs the excess of the mean of the interval, $(T_1+T_2)/2 - T_s$, not $T_1 - T_s$. Writing $80 - 20$ instead of $75 - 20$ corrupts every number that follows.

Applying the law at large excess. It is the small-excess linearisation of $P = e\sigma A(T^4 - T_s^4)$ — good to about 5% for a 10 K excess above a 300 K room, wrong by 38% at 100 K. JEE-Advanced punishes this: a glowing body needs Stefan, not Newton. (Note the kelvin rule too: differences may stay in °C, but the moment you use $T^4$, absolute kelvin is compulsory.)

Which board wants what. NCERT Class 11 and the published JEE-Advanced syllabus keep Newton's law of cooling by name, and it is standard in ICSE, CBSE, NEET and A-Level. One wrinkle: the 2024 NTA rationalisation dropped the literal phrase from the JEE-Main syllabus PDF, though heat transfer remains in Unit 7. Coaching material and previous-year JEE-Main papers still use it freely, so learn it regardless.

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) — the statement, the exponential and the standard two-interval numericals · ISC/ICSE Concise Physics, Heat — the calorimeter cooling-curve experiment · H. C. Verma, Concepts of Physics Vol. 2, Heat Transfer.

Provenance. Standard results; every number is recomputed in the verifier — the average-temperature anchor ($\Delta t_2 = 6.11$ min, $k = 0.0364\ \mathrm{min^{-1}}$) and its exact-exponential cross-check ($0.0365\ \mathrm{min^{-1}}$, $6.12$ min); the coffee exponential ($k = 0.0486\ \mathrm{min^{-1}}$, $T(20) = 49.6\ \mathrm{^\circ C}$, $T(30) = 40.1\ \mathrm{^\circ C}$, $30.2$ min to 40 °C, $t_{1/2} = 14.3$ min, $\tau = 20.6$ min, factor $0.615$, drops 25.0/15.4/9.5 °C, rates 3.2 and 0.45 °C/min); the board answer $300/7 = 42.9\ \mathrm{^\circ C}$; $T_s = 16\ \mathrm{^\circ C}$ ($k = 0.0571$ vs $0.0575\ \mathrm{min^{-1}}$); the lab slope $0.05\ \mathrm{min^{-1}}$ and $1.2\ \mathrm{^\circ C/min}$ at excess 24 °C; the $hA/(mc)$ estimate $0.029\ \mathrm{min^{-1}}$; the linearisation errors (5% at 10 K, 38% at 100 K); the forensic 5.3 h.

Prerequisites. Conduction & Thermal Conductivity · Thermal Radiation & the Black Body. Leads to. ARCHETYPE: Cooling & Radiation Numericals (where the full $T^4$ machinery takes over). Siblings. Convection (the origin of $h$) · Stefan–Boltzmann & Wien's Laws (the exact law this one approximates). Go deeper (UG). "Stefan-Boltzmann Law and Radiation Pressure" treats the radiation side exactly, and "Transport Coefficients and Green–Kubo Relations" explains where a coefficient like $h$ comes from microscopically — enrichment pointers only.