Thermal Properties of Matter · Lesson 7 of 8
Newton’s Law of Cooling
“Derive and test Newton’s small-temperature-difference cooling law, interpret its exponential solution and distinguish exact model predictions from interval-average estimates.”
• State the cooling law with correct signs, units and assumptions. • Derive the exponential temperature-excess decay by separation and integration. • Use excess-temperature and logarithmic graphs to test a controlled experiment. • Compare interval-average and exact exponential cooling-time calculations.
A hot sample cools fastest when its excess temperature is largest
Put warm water in a calorimeter and record its temperature at equal time intervals after the heater is removed. Its temperature usually falls rapidly at first, then more slowly as it approaches room temperature. The relevant quantity is the excess temperature θ = T − Ts, where T is the sample temperature and Ts is its surroundings’ temperature. A drop from 70 °C to 60 °C is not the same thermal driving condition in a 20 °C room as in a 50 °C room.
A useful investigation uses about 300 mL of water initially around 40 °C above room temperature. Gently stir and record both sample and surrounding temperatures at equal intervals, continuing until the sample is only about 5 °C above its surroundings. Plot excess temperature against time and estimate local slopes using nearby measurements. The experiment is an opportunity to test a model: the higher initial difference may not satisfy the small-difference approximation as closely as the later data.
Separate energy-loss rate from temperature-loss rate
The derivative dT/dt means the instantaneous temperature change per unit time: graphically it is the tangent slope on a temperature-versus-time curve. Multiplying by mc gives the rate of change of the body’s internal energy in the single-state, constant-capacity model. Outward loss is positive by our chosen convention, so the body’s energy rate is the negative of that loss. This avoids confusing a negative cooling slope with a negative heat-transfer coefficient.
The model requires approximately uniform sample temperature, constant surroundings, approximately constant heat capacity and a heat-transfer coefficient that stays reasonably stable. Stirring helps the uniform-temperature assumption for a liquid. A larger heat capacity slows its temperature change for the same surface transfer conditions. A different mass, surface area, airflow or vessel can change κ, so it cannot automatically be carried from one apparatus to another.
The law is especially useful for small temperature differences. Conduction, convection and radiation can contribute to the effective coefficient. For radiation near a fixed absolute Ts, write T = Ts + θ and expand T⁴ − Ts⁴: the first term is 4Ts³θ, while higher powers of θ are small when |θ|/Ts is small. This explains why a linear difference law can approximate an underlying fourth-power radiative law. It does not make the exact radiation law linear for arbitrarily large differences.
Derive the exponential approach to equilibrium
Assume Ts is constant, so dθ/dt = dT/dt. Move the changing θ to one side: dθ/θ = −κ dt. Integrating sums these small fractional changes through the cooling interval. Between initial excess θ0 at time zero and excess θ at time t, the integral of 1/θ is a logarithmic ratio. Writing the ratio avoids taking a logarithm of a dimensional quantity.
Exponentiating the logarithmic equation means using exp(ln x) = x, giving θ/θ0 = exp(−κt). Equal time intervals reduce the excess by the same factor, rather than subtracting the same number of degrees. The ideal curve approaches Ts asymptotically; after a finite time its excess can be experimentally negligible even though the exact exponential has not reached zero. The same signed equation describes warming toward a hotter surrounding temperature when its assumptions hold.
Verify the law with a stable surrounding bath
Place the warm-water calorimeter in a surrounding vessel whose water temperature is monitored separately. Keep the surrounding bath large or controlled, stir gently to reduce internal gradients, and record both temperatures at equal time steps. Compute the sample’s excess above the measured bath and plot ln(θ/θ0) against time. A straight line with negative slope supports a constant κ over that interval. Curvature can reveal changing bath temperature, changing transfer conditions or a difference too large for the approximation. Random scatter alone does not establish a new law.
A short-interval shortcut and its limitation
Over a sufficiently small temperature interval, replace the changing temperature by its interval average, Tavg = (Ti + Tf)/2. Then approximate the average cooling rate as (Ti − Tf)/Δt ≈ κ(Tavg − Ts). This treats the current as nearly constant within that interval. It is convenient for quick estimates, but it is not algebraically identical to the exponential solution. Its accuracy improves when the excess changes by only a small fraction within the interval.
Problem
A pan cools from 94 °C to 86 °C in 2 min in a 20 °C room. Estimate the time from 71 °C to 69 °C under unchanged transfer conditions.
- 1.The first interval’s average temperature is 90 °C, giving an average excess of 70 K and average cooling rate 8/2 = 4 K min⁻¹.
- 2.The average-interval estimate is κ ≈ 4/70 min⁻¹. The second average excess is 70 − 20 = 50 K, so time ≈ 2/[(4/70) × 50] = 0.700 min = 42.0 s.
- 3.Using the exponential law instead, κ = ln(74/66)/2 = 0.057205 min⁻¹.
- 4.The exact time within that constant-κ model is ln(51/49)/κ = 0.69933 min ≈ 41.96 s.
- 5.Both methods round to about 42 s here. Their agreement follows from the modest fractional excess changes, not from the two methods being identical.
Problem
A body cools from 80 °C to 50 °C in 5 min while its surroundings remain at 20 °C. Find the later time from 60 °C to 30 °C, assuming one constant κ.
- 1.The first interval changes excess from 60 K to 30 K: it is one halving in 5 min, so κ = ln 2/5 min⁻¹.
- 2.The later excess changes from 40 K to 10 K, a factor of four or two halvings. The exponential prediction is therefore 10 min.
- 3.The interval-average shortcut instead estimates κ ≈ (30/5)/45 = 0.13333 min⁻¹ from the first interval.
- 4.Using the later average excess of 25 K gives 30/(0.13333 × 25) = 9 min. This is the interval-average estimate, whereas 10 min is exact within the stated constant-κ exponential model.
- 5.The later interval changes excess substantially, making the approximation difference noticeable. Real-world model error remains separate from this difference between solution methods.
Quick Quiz
Quiz
In Newton cooling, which difference drives the modelled outward current?
The driving quantity is excess temperature above the surroundings.
For a warmer body cooling passively, the sign of dT/dt is
The body temperature decreases while κ and its positive excess remain positive.
If excess temperature halves in 4 min with constant κ, a reduction to one quarter takes
The exponential gives the same fractional reduction in equal intervals: two halvings take 8 min.
A straight line is expected when plotting
The integrated logarithmic ratio equals −κt.
Why can the average-temperature shortcut disagree with an exact exponential calculation?
Large fractional excess changes can make the averaging error noticeable.
Practice Problems
- A sample in a 20 °C room cools from 60 °C to 40 °C in 6.0 min. Find κ in min⁻¹ and the later time from 30 °C to 25 °C using the constant-κ exponential model.
- For T0 = 70 °C, Ts = 25 °C and κ = 0.0500 min⁻¹, find the temperature after 10.0 min. Explain why subtracting a fixed number of degrees each minute would be incorrect.
- A sample of mass 0.200 kg and specific heat 400 J kg⁻¹ K⁻¹ has kc = 0.800 W K⁻¹. Find κ and the initial outward power and dT/dt when T − Ts = 15 K. State the sign convention.
- The measured excess temperatures at 0, 2, 4 and 6 min are 40, 32, 25.6 and 20.48 K. Test whether these data fit one constant κ. Describe the corresponding logarithmic-ratio plot.
- For cooling from 80 °C to 50 °C in 5 min at ambient 20 °C, compare the exact and interval-average estimates for the later 60 °C to 30 °C interval. Explain the source of the difference and give two experimental conditions that could invalidate both calculations.
Key Takeaways
• A cooling slope is negative for a warm body, while its outward energy-loss current is positive. • Constant surroundings and transfer conditions lead to exponential excess-temperature decay, with κ = kc/(mc). • Equal times give equal fractional excess reductions; a logarithmic ratio gives a straight-line test. • An interval-average shortcut is an approximation, and the cooling law itself also has physical limits.