Newton’s Law of Cooling

Newton’s law of cooling models the transient temperature of a body exchanging heat with a surrounding fluid by convection. When internal temperature gradients inside the body are negligible (the lumped-capacitance assumption), the body’s temperature decays (or rises) exponentially toward the ambient.

Statement and Governing Equation

The convective heat transfer rate is proportional to the instantaneous temperature difference between a surface and the surrounding fluid:

\[ \dot{Q}_{\text{conv}} = h\,A\,(T_s - T_\infty) \]

For a lumped body where the internal temperature is spatially uniform and equal to \(T(t)\), an energy balance gives:

\[ m\,c\,\frac{dT}{dt} = -\,h\,A\,\big(T - T_\infty\big) \quad\Longrightarrow\quad \frac{dT}{dt} = -\frac{1}{\tau}\,\big(T - T_\infty\big) \]

with the thermal time constant

\[ \tau = \frac{m\,c}{h\,A} = \frac{\rho\,c\,V}{h\,A} \]

Solution Forms

Constant Ambient Temperature

For constant \(T_\infty\) and initial condition \(T(0)=T_0\):

\[ T(t) = T_\infty + \big(T_0 - T_\infty\big)\,e^{-t/\tau} \]

Time-Varying Ambient Temperature

If \(T_\infty = T_\infty(t)\) varies in time, the ODE is linear and solved via an integrating factor:

\[ \frac{dT}{dt} + \frac{1}{\tau}\,T = \frac{1}{\tau}\,T_\infty(t) \quad\Rightarrow\quad T(t) = e^{-t/\tau}\,T_0 + \frac{1}{\tau}\,e^{-t/\tau}\!\int_{0}^{t} e^{s/\tau}\,T_\infty(s)\,ds \]

With Constant Internal Heat Generation

If the body generates heat uniformly at rate \(\dot{Q}_{\text{gen}}\) (W):

\[ m\,c\,\frac{dT}{dt} = -hA\,(T - T_\infty) + \dot{Q}_{\text{gen}} \quad\Rightarrow\quad T(t) = T_\infty + \Delta T_{\text{ss}} + \big(T_0 - T_\infty - \Delta T_{\text{ss}}\big)e^{-t/\tau} \]

where the steady temperature rise is

\[ \Delta T_{\text{ss}} = \frac{\dot{Q}_{\text{gen}}}{hA} \]

Validity: Lumped-Capacitance Criterion

Newton’s law (exponential model) requires negligible internal temperature gradients. The standard check uses the Biot number:

\[ \mathrm{Bi} = \frac{h\,L_c}{k} \;\lesssim\; 0.1 \quad\text{with}\quad L_c = \frac{V}{A} \]

• Material conductivity: \(k\) is the solid’s thermal conductivity.
• Characteristic length examples: slab of thickness \(2L\): \(L_c=L\) (one side) or \(L/2\) (both sides); long cylinder: \(L_c \approx r/2\) (lateral cooling); sphere: \(L_c = r/3\).
• If Bi > 0.1: Internal gradients matter; use spatial solutions (e.g., Heisler charts/eigenfunction expansions) instead of lumped model.

Including Radiation and Other Effects

Combined Convection and Radiation

Radiation can be linearized about a mean temperature \(T_m\) to define an effective radiative coefficient:

\[ h_{\text{rad}} \approx 4\,\varepsilon\,\sigma\,T_m^{3} \quad\Rightarrow\quad h_{\text{tot}} \approx h_{\text{conv}} + h_{\text{rad}},\quad \tau \approx \frac{m c}{A\,h_{\text{tot}}} \]

• \(\varepsilon\): surface emissivity; \(\sigma\): Stefan–Boltzmann constant.
• Use \(T_m \approx (T + T_\infty)/2\) for a first estimate; iterate if needed.

Time-Dependent Coefficients

If \(h\), \(A\), or \(c\) vary with time, the ODE becomes

\[ \frac{dT}{dt} + \frac{h(t)A(t)}{m(t)c(t)}\big(T - T_\infty(t)\big) = 0 \]

The formal solution is

\[ T(t) = e^{-\int_0^t \frac{ds}{\tau(s)}}\,T_0 + e^{-\int_0^t \frac{ds}{\tau(s)}} \int_{0}^{t} \frac{T_\infty(u)}{\tau(u)}\,e^{\int_0^u \frac{ds}{\tau(s)}}\,du \quad\text{where}\quad \tau(t)=\frac{m(t)c(t)}{h(t)A(t)} \]

Estimating h from Cooling Curves

Rearranging the constant-ambient solution:

\[ \ln\!\big(T(t)-T_\infty\big) = \ln\!\big(T_0 - T_\infty\big) - \frac{t}{\tau} \quad\Rightarrow\quad \text{slope} = -\frac{1}{\tau} = -\frac{hA}{m c} \]

• Plot \(\ln(T-T_\infty)\) vs time; the slope yields \(h\) if \(m,c,A\) are known.
• Use the linear region where Bi remains small and properties are approximately constant.

Worked Examples

Example 1 — Aluminum Sphere Cooling in Air (Lumped)

• Data: Sphere radius \(r=10\,\text{mm}\), \(\rho=2700\,\text{kg/m}^3\), \(c=900\,\text{J/(kg·K)}\), \(k=205\,\text{W/(m·K)}\), \(h=30\,\text{W/(m}^2\text{·K)}\), \(T_0=100^\circ\text{C}\), \(T_\infty=25^\circ\text{C}\).
• Geometry: \(V=\tfrac{4}{3}\pi r^3=4.1888\times10^{-6}\,\text{m}^3\); \(A=4\pi r^2=1.2566\times10^{-3}\,\text{m}^2\); \(m=\rho V=0.01131\,\text{kg}\).
• Biot number: \(L_c=V/A=r/3=3.33\,\text{mm}\Rightarrow \mathrm{Bi}=hL_c/k \approx 30\cdot 0.00333/205 \approx 4.9\times10^{-4} \ll 0.1\) (lumped valid).
• Time constant: \(\tau=\dfrac{m c}{hA}=\dfrac{0.01131\cdot 900}{30\cdot 0.0012566}\approx 270\,\text{s}\).
• Temperature after 5 min: \(t=300\,\text{s}\Rightarrow T(t)=T_\infty+(T_0-T_\infty)e^{-t/\tau}=25+(75)\,e^{-300/270}\approx 25+24.7=49.7^\circ\text{C}\).

Example 2 — Adding Radiation to Air Cooling

• Data: Same sphere; \(\varepsilon=0.8\), take \(T_m\approx 350\,\text{K}\) (midway between object and ambient early on).
• Radiative coefficient: \(h_{\text{rad}}\approx 4\varepsilon\sigma T_m^3 \approx 4\cdot 0.8 \cdot 5.6704\times10^{-8}\cdot 350^3 \approx 7.8\,\text{W/(m}^2\text{·K)}\).
• Effective coefficient: \(h_{\text{tot}}\approx 30+7.8=37.8\,\text{W/(m}^2\text{·K)}\Rightarrow \tau \approx \dfrac{m c}{A h_{\text{tot}}}\approx 270\cdot \dfrac{30}{37.8}\approx 214\,\text{s}\).
• The body cools faster due to radiation; update \(T_m\) as \(T\) evolves for improved accuracy.

Example 3 — Internal Generation in a Small Device

• Data: Same sphere with \(\dot{Q}_{\text{gen}}=2\,\text{W}\); \(h=30\,\text{W/(m}^2\text{·K)}\); ambient \(T_\infty=25^\circ\text{C}\).
• Steady rise: \(\Delta T_{\text{ss}}=\dfrac{\dot{Q}_{\text{gen}}}{hA}=\dfrac{2}{30\cdot 0.0012566}\approx 53\,\text{K}\Rightarrow T_{\text{ss}}\approx 78^\circ\text{C}\).
• Transient: \(T(t)=T_\infty+\Delta T_{\text{ss}}+\big(T_0-T_\infty-\Delta T_{\text{ss}}\big)e^{-t/\tau}\).

Example 4 — Estimating h from Data

• Data: Measured \(T(t)\) for a small copper cylinder; \(m c\) and \(A\) known; \(T_\infty\) constant.
• Method: Compute \(Y(t)=\ln(T(t)-T_\infty)\). Fit a straight line \(Y=Y_0 - t/\tau\); slope \(= -1/\tau\).
• Result: \(h = \dfrac{m\,c}{A\,\tau}\). Validate by checking \(\mathrm{Bi}\ll 0.1\) using estimated \(h\).

Practical Guidance

• Check Biot first: Use \(\mathrm{Bi} \le 0.1\) as a rule of thumb for lumped validity.
• Property consistency: If \(c\) or \(h\) varies strongly with \(T\), segment time or iterate \(\tau\) over temperature bands.
• Geometry awareness: Use the correct \(L_c=V/A\) for \(\mathrm{Bi}\); for composite bodies, estimate an effective \(L_c\) or verify with a more detailed model.
• Combined modes: Radiation may be significant at elevated temperatures; include \(h_{\text{rad}}\).
• Data fitting: Use the early linear region of \(\ln(T-T_\infty)\) vs \(t\); late-time noise and drifting \(T_\infty\) can bias estimates.

Quick Reference

• Newton’s law (lumped): \[ m c \frac{dT}{dt} = -hA\,(T - T_\infty) \quad\Rightarrow\quad T(t)=T_\infty + (T_0 - T_\infty)e^{-t/\tau},\;\; \tau=\frac{m c}{hA} \]
• Biot number: \[ \mathrm{Bi}=\frac{h L_c}{k}\;\;(\text{require }\mathrm{Bi}\lesssim 0.1),\quad L_c=\frac{V}{A} ]
• With radiation: \[ h_{\text{rad}}\approx 4\varepsilon\sigma T_m^3,\quad h_{\text{tot}}\approx h_{\text{conv}}+h_{\text{rad}} \]
• Internal generation: \[ T(t)=T_\infty+\frac{\dot{Q}_{\text{gen}}}{hA}+\Big(T_0-T_\infty-\frac{\dot{Q}_{\text{gen}}}{hA}\Big)e^{-t/\tau} \]
• Estimating h: \[ \ln(T-T_\infty)=\ln(T_0-T_\infty)-\frac{t}{\tau},\quad h=\frac{m c}{A\,\tau} \]