Newton's Law of Cooling

Newton's hypothesis

Isaac Newton observed that a body cools (or warms) at a rate proportional to its temperature difference from the surroundings: $\boxed{\, \frac{dT}{dt} = -k \, (T - T_{\text{amb}}) \,}$

where:

• $T(t)$ is the temperature of the body at time $t$,
• $T_{\text{amb}}$ is the ambient (room) temperature,
• $k > 0$ is a positive constant depending on the object's material, surface area, and the heat-transfer medium.

The bigger the gap $T - T_{\text{amb}}$, the faster the heat flows (or the faster the body warms if the gap is negative). As the gap closes, the cooling slows — asymptotically reaching $T_{\text{amb}}$.

This is a first-order linear ODE (#174), identical in structure to "$y' + P y = Q$" with $P = k$, $Q = k \, T_{\text{amb}}$.

The given problem

• Initial coffee temperature: $T(0) = 90$ °C.
• Ambient (room): $T_{\text{amb}} = 20$ °C.
• Cooling constant: $k$ (adjustable via slider).

Step-by-step

Step 1 — Recognise the structure.

$\dfrac{dT}{dt} = -k (T - T_{\text{amb}})$ is first-order linear. Equilibrium is the value where $dT/dt = 0$, namely $T = T_{\text{amb}}$.

Step 2 — General solution by separation or by linear-ODE formula.

Separating variables: $\frac{dT}{T - T_{\text{amb}}} = -k \, dt$ $\ln|T - T_{\text{amb}}| = -k t + C$ $T - T_{\text{amb}} = A \, e^{-k t} \qquad (A = \pm e^{C})$ $T(t) = T_{\text{amb}} + A \, e^{-k t}.$

Step 3 — Apply initial condition.

$T(0) = T_{\text{amb}} + A = 90 \implies A = 90 - T_{\text{amb}} = 70$.

$\boxed{\, T(t) = 20 + 70 \, e^{-k t} \,}$

Interpretation — two terms

• Ambient $T_{\text{amb}} = 20$ °C: the "floor" the temperature asymptotes to.
• Transient $70 \, e^{-k t}$: the initial gap decaying exponentially.

The coffee is "initially 70°C above ambient"; that excess dies off at rate $k$.

Time constant

$\tau = 1/k$ is the time constant. After one $\tau$, the gap has decayed to $1/e \approx 37\%$ of its initial value: $T(\tau) - T_{\text{amb}} = 70 \cdot e^{-1} \approx 25.8 \text{ °C gap}$ $T(\tau) \approx 20 + 25.8 = 45.8 \text{ °C}.$

Useful rules of thumb:

• After $\tau$: ~63% of the way to ambient.
• After $2\tau$: ~86%.
• After $3\tau$: ~95%.
• After $5\tau$: ~99.3%.

For a real coffee cup with $k \approx 0.02$ min⁻¹ (so $\tau = 50$ min), it takes about 2.5 hours to essentially reach room temperature.

Verification

$T(t) = 20 + 70 e^{-kt}$. $T'(t) = -70 k e^{-kt}$. $-k(T - T_{\text{amb}}) = -k(70 e^{-kt}) = -70 k e^{-kt}$. ✓

Warming also follows the same law

If the body is colder than ambient, the gap is negative and the exponential decay still applies — the object warms up to $T_{\text{amb}}$. Same equation, different sign of $A$:

• Iced drink at 4°C in a 22°C room: $T(t) = 22 - 18 e^{-k t}$.

One formula for both cooling and warming. That's the power of linearity.

What determines $k$?

$k$ lumps several physics into one parameter: $k = \frac{h \cdot A}{m \cdot c_p}$

where

• $h$ is the heat transfer coefficient (W/m²·K) of the surface-to-air interface,
• $A$ is the surface area (m²),
• $m$ is the mass (kg),
• $c_p$ is the specific heat capacity (J/kg·K).

Takeaways:

• Bigger surface area ⇒ faster cooling (why blowing on soup helps — increases effective $h$ via convection and spreads out the soup).
• Larger mass or higher specific heat ⇒ slower cooling (why a big pot keeps tea warmer than a small cup).
• Vacuum insulation (thermos) drops $h$ dramatically, so $k$ becomes tiny and $\tau$ huge.

Limitations of Newton's law

Newton's law is a linearisation. It works well when $T - T_{\text{amb}}$ is small compared to absolute temperatures. For very hot objects, radiative heat loss dominates and obeys the Stefan-Boltzmann law: flux $\propto T^4 - T_{\text{amb}}^4$. A pot of lava at 1200 K in a 300 K room cools much faster at first than Newton predicts.

For mildly warm everyday objects (drinks, bodies, electronics), Newton's linear approximation is very good.

The forensic / medical version

Forensic investigators use Newton's law (often in refined form) to estimate time of death: given body temperature and ambient temperature at the discovery scene, back-solve for $t$. With a typical human $k \approx 0.1$ hr⁻¹ (so $\tau \approx 10$ hr), the formula $T(t) = 20 + (37 - 20) \, e^{-kt}$ lets an investigator approximate time since death to within an hour or two — assuming nothing else messes with the body's thermodynamics.

Common mistakes

• Sign of $k$. The law is $dT/dt = -k(T - T_{\text{amb}})$, with $k > 0$. Dropping the minus sign gives an exponential blow-up, which is wrong.
• Using $T$ instead of $T - T_{\text{amb}}$. The rate depends on the difference from ambient, not the absolute temperature.
• Assuming constant $T_{\text{amb}}$. A coffee left outside on a cold night will chase a moving ambient; the ODE then has a time-dependent right-hand side.
• Ignoring radiation for very hot objects. Newton's law breaks down when the temperature difference is large.

Try it in the visualization

Plot the cooling curve $T(t) = T_{\text{amb}} + (T_0 - T_{\text{amb}}) e^{-kt}$ with the horizontal asymptote at $T_{\text{amb}}$. Mark the time constant $\tau = 1/k$ where the curve has decayed to 37% of its initial gap. Slide $k$ to see the curve steepen or flatten; slide $T_{\text{amb}}$ to shift the asymptote.