Physics / Newton's Law of Cooling
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EXPERIMENT 25 · PHYSICS LABORATORY

Newton's Law of Cooling theory

Understand newton's law of cooling: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
ln[(T_i − T_0) / (T − T_0)] = k · t ⇔ Y = k · X ⇔ k = S

Keep the physical quantities, graph variables and units consistent.

01Newton's Empirical Law of Cooling: Formulation and Physical Regime

In 1701, Sir Isaac Newton published his landmark thermal treatise Scala Graduum Caloris ('Scale of the Degrees of Heat') in the Philosophical Transactions of the Royal Society. Newton sought an objective method to calibrate high-temperature thermometers beyond the boiling point of water by observing the rate at which glowing iron ingots cooled in an air stream. He observed that the quantity of heat lost by a body in a given small increment of time is proportional to the difference between the temperature of the body and that of the surrounding medium: dQ/dt = −h_eff A (T − T_0), where h_eff is the effective overall heat transfer coefficient, A is the surface area of the body, and T_0 is the ambient enclosure temperature. Substituting the caloric relation dQ = m c dT (where m is the mass and c is the specific heat capacity), the governing rate equation becomes dT/dt = −[h_eff A / (m c)] (T − T_0) = −k (T − T_0). The proportionality factor k, termed the cooling constant, depends intrinsically on the geometric shape and surface area of the vessel, the surface emissivity, the heat capacity of the contents, and the velocity of convective currents. Newton's formulation is an empirical approximation that holds with remarkable precision for moderate temperature differences (typically θ = T − T_0 < 30°C to 40°C), where nonlinear radiative and turbulent convective terms can be legitimately linearized.

\frac{dT}{dt} = -k (T - T_0) \quad \text{where } k = \frac{h_{\text{eff}} A}{m c}
Under what physical conditions does Newton's law of cooling provide an accurate description of thermal dissipation?

Newton's law of cooling is valid when the temperature excess (T − T_0) between the cooling body and its surroundings is relatively small (typically less than 30°C to 40°C), heat loss occurs in a constant-temperature enclosure free from forced drafts, and natural convection currents remain laminar while radiative emission can be linearized.

02Radiative Heat Loss and Stefan-Boltzmann Linearization

Thermal radiation from any physical surface is fundamentally governed by the Stefan-Boltzmann law, established empirically by Jožef Stefan in 1879 and derived thermodynamically by Ludwig Boltzmann in 1884. The net radiative power P_rad exchanged between a body of surface area A, absolute temperature T (in kelvins), and surface emissivity ε, situated inside an isothermal enclosure at absolute temperature T_0, is given by P_rad = ε σ A (T⁴ − T_0⁴), where σ = 5.670374 × 10⁻⁸ W m⁻² K⁻⁴ is the Stefan-Boltzmann constant. Although this fourth-power dependence appears starkly non-linear, it can be factored algebraically: T⁴ − T_0⁴ = (T − T_0)(T + T_0)(T² + T_0²). When the temperature excess ΔT = T − T_0 is small compared to the absolute temperature T_0 (such that ΔT / T_0 ≪ 1), we approximate T ≈ T_0, yielding (T + T_0) ≈ 2 T_0 and (T² + T_0²) ≈ 2 T_0². Consequently, the factor (T⁴ − T_0⁴) simplifies to 4 T_0³ (T − T_0). The radiative power loss becomes P_rad ≈ [4 ε σ A T_0³] (T − T_0) = h_rad A (T − T_0), where h_rad = 4 ε σ T_0³ is the linearized radiative heat transfer coefficient. At room temperature (T_0 = 293.15 K), h_rad evaluates to approximately 5.7 ε W m⁻² K⁻¹. Thus, thermal radiation adheres strictly to Newton's linear proportionality at modest temperature excesses.

T^4 - T_0^4 \approx 4 T_0^3 (T - T_0) \implies P_{\text{rad}} \approx 4 \varepsilon \sigma A T_0^3 (T - T_0)
How does the Stefan-Boltzmann fourth-power radiation law reduce to Newton's linear cooling law?

By factoring (T⁴ − T_0⁴) = (T − T_0)(T + T_0)(T² + T_0²) and expanding in powers of (T − T_0)/T_0, for small temperature differences (T ≈ T_0) the last two terms approach 2T_0 and 2T_0² respectively, giving T⁴ − T_0⁴ ≈ 4 T_0³ (T − T_0). Net radiative loss is therefore directly proportional to (T − T_0).

03Natural Convection and Boundary Layer Transport

In addition to radiation, a cooling body immersed in air or a fluid transfers thermal energy through natural (free) convection. As heat conducts from the outer calorimeter wall into the adjacent fluid layer, the fluid warms, expands, and undergoes a localized decrease in mass density. Under the buoyant influence of gravity, the warmer, lighter fluid ascends along the vertical boundaries of the vessel, while cooler, denser fluid descends to take its place, establishing a recirculating convective boundary layer. The rate of convective heat transfer is conventionally modeled through Nusselt numbers and Rayleigh numbers: q_conv = h_conv A (T − T_0)^n. For laminar free convection over vertical cylinders, boundary-layer theory reveals that the exponent n is approximately 5/4 = 1.25, rather than exactly 1.0. However, over moderate temperature spans of 10°C to 30°C, the variation of (T − T_0)^0.25 is sufficiently modest that an effective, averaged convective coefficient h_conv can be defined. Combining linearized radiation and natural convection yields the composite heat transfer coefficient h_eff = h_rad + h_conv. Suppressing turbulent air drafts by enclosing the calorimeter in a stagnant double-walled jacket ensures that the convective boundary layer remains steady, laminar, and highly reproducible.

q_{\text{conv}} = h_{\text{conv}} A (T - T_0)^{1.25} \approx \bar{h}_{\text{conv}} A (T - T_0)
Why is the rate of heat loss by natural convection slightly superlinear with temperature difference?

In natural convection, the buoyant driving force that accelerates the fluid is itself proportional to the temperature difference (T − T_0). Higher temperature excesses simultaneously increase both the thermal gradient and the convective circulation velocity, producing an empirical power law of approximately (T − T_0)^1.25.

04Differential Formulation and Exponential Temperature Decay

The mathematical foundation of cooling kinetics rests upon the separation of variables in the linear differential equation dT/dt = −k (T − T_0). Defining the temperature excess θ(t) = T(t) − T_0 and differentiating yields dθ/dt = dT/dt, since the enclosure temperature T_0 is constant. The differential equation simplifies to dθ/dt = −k θ, or equivalently dθ / θ = −k dt. Integrating both sides from t = 0 (where θ = θ_0 = T_i − T_0) to elapsed time t yields ln(θ / θ_0) = −k t, which exponentiates to the classical exponential decay law: θ(t) = θ_0 exp(−k t). The temperature of the cooling body as a function of time is therefore T(t) = T_0 + (T_i − T_0) exp(−k t). The characteristic thermal time constant of the system is defined as τ = 1/k. Physically, τ represents the time required for the temperature excess to decrease to 1/e (approximately 36.787%) of its initial value θ_0. For our experimental calorimeter system with k = 0.0960 min⁻¹, the thermal time constant is τ = 10.42 min (625.0 s). Half-life decay time occurs at t_half = τ ln(2) = 0.69315 / 0.0960 ≈ 7.22 min.

\theta(t) = \theta_0 e^{-k t} \iff \ln\left(\frac{\theta_0}{\theta(t)}\right) = k t \quad \text{where } \tau = \frac{1}{k}
What is the physical significance of the thermal time constant τ = 1/k in cooling phenomena?

The thermal time constant τ represents the time required for the temperature excess above ambient to decay to 1/e (approx. 36.8%) of its initial value. It characterizes the thermal inertia of the system: systems with larger heat capacities or smaller surface areas exhibit larger time constants and cool more slowly.

05The Method of Cooling for Comparative Specific Heat Determinations

Newton's cooling law provides an elegant historical method—extensively utilized by Dulong, Petit, and Regnault—to determine the specific heat capacity of an unknown liquid without requiring direct heat addition or mechanical mixing. Consider two identical calorimeter vessels possessing identical outer geometries, surface finishes, and surface areas A, suspended within the same isothermal water jacket. One vessel is filled with a mass m_w of water (specific heat c_w), while the second is filled with an equal volume of test liquid of mass m_L (unknown specific heat c_L). Because the radiating surfaces and enclosure temperatures are identical, both vessels lose heat at the identical absolute rate dQ/dt when passing through the same temperature range from T_a to T_b. The total heat lost by the water vessel during time interval Δt_w is ΔQ = (m_w c_w + W_cal) (T_a − T_b), while for the test liquid during interval Δt_L it is ΔQ = (m_L c_L + W_cal) (T_a − T_b). Dividing the two heat loss expressions gives (m_w c_w + W_cal) / Δt_w = (m_L c_L + W_cal) / Δt_L. Solving directly yields the specific heat c_L = [(m_w c_w + W_cal)(Δt_L / Δt_w) − W_cal] / m_L. This comparative method eliminates absolute power calibrations by converting thermal metrology into precision timing measurements.

\frac{m_w c_w + W_{\text{cal}}}{\Delta t_w} = \frac{m_L c_L + W_{\text{cal}}}{\Delta t_L} \implies c_L = \frac{(m_w c_w + W_{\text{cal}})\frac{\Delta t_L}{\Delta t_w} - W_{\text{cal}}}{m_L}
How does the method of cooling allow the specific heat of a liquid to be determined by measuring cooling times?

Because two identical vessels with identical surface finishes lose heat at the same rate when at the same temperature, the heat lost over a given temperature interval is directly proportional to the time required to cool through that interval. Equating the heat loss rates allows the unknown specific heat to be found from the ratio of cooling times.

06Surface Emissivity, Polishing, and Enclosure Geometry

The cooling constant k is strongly modulated by the optical surface state of the radiating vessel. According to Kirchhoff's law of thermal radiation, for an arbitrary body in thermodynamic equilibrium with radiation, the monochromatic emissivity ε(λ, T) equals the absorptivity α(λ, T). A dull, soot-blackened copper vessel exhibits an emissivity close to unity (ε ≈ 0.95 to 0.98), functioning nearly as an ideal blackbody radiator. In contrast, a freshly polished, specular copper surface exhibits an emissivity of ε ≈ 0.03 to 0.05 in the infrared spectrum (wavelengths 8 μm to 14 μm corresponding to thermal emission at 300 K to 350 K). Consequently, a blackened calorimeter loses thermal radiation at a rate roughly twenty to thirty times greater than an identically shaped polished vessel, resulting in a substantially higher cooling constant k_black ≫ k_polished. In our laboratory configuration, maintaining a polished exterior minimizes radiation, isolating conductive and convective transport across the stagnant annular air gap. Furthermore, the enclosure geometry—characterized by a large ratio of jacket volume to calorimeter volume—ensures that radiated photons are absorbed by the jacket without significantly raising the sink temperature T_0.

P_{\text{rad}} = \varepsilon \sigma A (T^4 - T_0^4) \implies k_{\text{black}} > k_{\text{polished}} \quad (\varepsilon_{\text{black}} \approx 0.95, \, \varepsilon_{\text{polished}} \approx 0.04)
Why does a blackened calorimeter cool significantly faster than an otherwise identical polished calorimeter?

A blackened surface possesses an emissivity close to unity (ε ≈ 0.95), radiating thermal energy nearly as an ideal blackbody. A polished copper surface reflects infrared radiation internally, exhibiting a low emissivity (ε ≈ 0.04). The radiative heat loss rate is proportional to emissivity, making the blackened vessel cool much faster.

07Thermometer Lag, Emergent Stem Corrections, and Non-Isothermal Gradients

Experimental verification of Newton's cooling law requires meticulous attention to thermometric errors. Three primary sources of experimental distortion must be controlled: thermometer thermal lag, emergent stem exposure, and internal fluid stratification. Because mercury-in-glass thermometer bulbs possess finite mass and thermal resistance, the recorded temperature lags the true bath temperature by a time constant τ_therm = m_bulb c_glass / (h_therm A_bulb). In a smoothly cooling bath, this produces an apparent time shift without altering the logarithmic slope k. However, if the thermometer is partially immersed, the mercury in the emergent stem residing above the calorimeter lid is cooler than the liquid bath, causing the indicated reading to fall below the true temperature. The emergent stem correction ΔT_stem = 0.00016 N (T − t_s) must be applied when high absolute accuracy is required. Most crucially, because water has a relatively low thermal conductivity, failure to stir continuously results in cold boundary layers hugging the container walls while the core remains warm, producing spurious non-exponential cooling trajectories. Gentle, periodic mechanical stirring guarantees spatial isothermality throughout the cooling cycle.

\Delta T_{\text{stem}} = 0.00016 \cdot N \cdot (T - t_s) \quad \text{and} \quad \tau_{\text{therm}} = \frac{m_{\text{bulb}} c_{\text{glass}}}{h_{\text{therm}} A_{\text{bulb}}}
Why is periodic mechanical stirring critical when measuring the cooling curve of water?

Water has low thermal conductivity, so without stirring, heat conducts slowly from the interior to the walls. Cooler water accumulates near the walls and surface, creating steep internal temperature gradients. Stirring enforces spatial isothermality, ensuring the thermometer accurately reflects the uniform bulk temperature of the cooling liquid.

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Connect the equations to a measurement.

The simulation evaluates water temperatures using the exact analytical solution T(t) = T_0 + (T_i − T_0) exp(−k t) with initial water temperature T_i = 80.0 °C, constant jacket temperature T_0 = 20.0 °C, and cooling constant k = 0.0960 min⁻¹ (time constant τ = 10.417 min). Readings are rounded to 0.01 °C to simulate high-precision calorimetric thermometry, and the linear transformation ln[(T_i − T_0) / (T − T_0)] = k t is evaluated using ordinary least-squares regression.