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Worked solution

Fit Newton cooling using the temperature excess

[Newton’s Law of Cooling]

The rate of temperature loss is proportional to the difference between a body and its surroundings. Express this as a differential equation, using t for time and theta for body temperature.

A rod is immersed in water held at 15°C. It cools from 60°C to 40°C in 3 minutes. Find (i) its temperature at time t, and (ii) its temperature immediately after 9 minutes.

CJC · Lecture · Example 9

State the signed rate law

If the surroundings have temperature T, dθ/dt=−k(θ−T)d\theta/dt=-k(\theta-T), k>0k>0. With T=15, dθ/dt=−k(θ−15)d\theta/dt=-k(\theta-15). The minus sign encodes cooling when the rod is warmer than its surroundings.

How did this part go?

(i) Separate, integrate and use the context

∫dθ/(θ−15)=∫−k dt\int d\theta/(\theta-15)=\int-k\,dt gives ln⁡∣θ−15∣=−kt+C\ln|\theta-15|=-kt+C. Here θ>15\theta>15, so θ=15+Ae−kt\theta=15+Ae^{-kt}. The teacher stresses using the physical range to remove the modulus and using one independent integration constant: two constants would combine into their difference.

How did this part go?

(i) Fit both given observations

At t=0t=0, 60=15+A60=15+A, hence A=45. At t=3t=3, 40=15+45e−3k40=15+45e^{-3k}, so e−3k=5/9e^{-3k}=5/9 and k=−13ln⁡(5/9)>0k=-\tfrac13\ln(5/9)>0. Thus θ(t)=15+45e(t/3)ln⁡(5/9)=15+45(5/9)t/3\theta(t)=15+45e^{(t/3)\ln(5/9)}=15+45(5/9)^{t/3} for t in minutes.

How did this part go?

(ii) Evaluate after nine minutes

θ(9)=15+45(5/9)3≈22.7∘C\theta(9)=15+45(5/9)^3\approx22.7^\circ\mathrm C (3 s.f.). It approaches 15°C from above as time increases.

How did this part go?

Common mistakes

Cooling is proportional to temperature excess, not necessarily to the body temperature itself.

Read the hint