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Lesson 99 — Newton's Law of Cooling

dT/dt = -k(T - T_amb): Separable ODE with exponential solution. Forensic, industrial, and everyday applications.

Used in: Spécialité Maths Terminale (França) · Leistungskurs Mathematik Klasse 12 (Alemanha) · H2 Mathematics (Singapura)

dTdt=k(TTamb)    T(t)=Tamb+(T0Tamb)ekt\frac{dT}{dt} = -k(T - T_{\text{amb}}) \;\Longrightarrow\; T(t) = T_{\text{amb}} + (T_0 - T_{\text{amb}})\,e^{-kt}
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Rigorous notation, full derivation, hypotheses

Rigorous derivation and solution

The law and its hypothesis

The rate of change of an object's temperature is proportional to the deviation from the environment:

dTdt=k(TTamb)\frac{dT}{dt} = -k(T - T_{\text{amb}})
what this means · k > 0 is the heat transfer constant [1/time]. The negative sign indicates that the object cools when T > T_amb and warms when T < T_amb.

"The temperature of a body changes at a rate proportional to the difference between the temperature of the body and the temperature of the surrounding medium. This is Newton's law of cooling." — Trench, Elementary Differential Equations §4.2

Time constant and half-life

Determining kk from data

Given T(t1)=T1T(t_1) = T_1:

k=1t1ln ⁣(T1TambT0Tamb)k = -\frac{1}{t_1}\ln\!\left(\frac{T_1 - T_{\text{amb}}}{T_0 - T_{\text{amb}}}\right)
what this means · k is isolated directly from a measurement at time t_1.

Model validity

Solved examples

Exercise list

30 exercises · 7 with worked solution (25%)

Application 17Understanding 3Modeling 7Challenge 1Proof 2
  1. Ex. 99.1Application

    T0=100T_0 = 100 °C, Tamb=22T_{\text{amb}} = 22 °C, k=0,04k = 0{,}04 min1^{-1}. Write T(t)T(t) and calculate T(15)T(15).

  2. Ex. 99.2Application

    T0=100T_0 = 100 °C, Tamb=20T_{\text{amb}} = 20 °C, T(10)=55T(10) = 55 °C. Determine kk.

  3. Ex. 99.3Application

    T0=5T_0 = 5 °C (cold object), Tamb=25T_{\text{amb}} = 25 °C, k=0,06k = 0{,}06 min1^{-1}. Calculate T(20)T(20).

  4. Ex. 99.4Application

    T0=90T_0 = 90 °C, Tamb=20T_{\text{amb}} = 20 °C, k=0,05k = 0{,}05 min1^{-1}. Calculate the half-life of the temperature difference and the temperature at that instant.

  5. Ex. 99.5ApplicationAnswer key

    T0=80T_0 = 80 °C, Tamb=10T_{\text{amb}} = -10 °C, k=0,03k = 0{,}03 min1^{-1}. Calculate τ\tau and T(τ)T(\tau).

  6. Ex. 99.6Application

    T0=100T_0 = 100 °C, Tamb=20T_{\text{amb}} = 20 °C, k=0,05k = 0{,}05 min1^{-1}. How long until T=40T = 40 °C?

  7. Ex. 99.7Application

    k=0,04k = 0{,}04 min1^{-1}. How long until the temperature difference drops to less than 1% of the initial value?

  8. Ex. 99.8Application

    Body found at 10:00 PM: T=33T = 33 °C. Tamb=18T_{\text{amb}} = 18 °C, Tnormal=37T_{\text{normal}} = 37 °C, k=0,06k = 0{,}06 h1^{-1}. Estimate the time of death.

  9. Ex. 99.9Application

    Container with liquid: h=20h = 20 W/(m²K), A=0,04A = 0{,}04 m², m=0,3m = 0{,}3 kg, cp=4000c_p = 4000 J/(kgK). Calculate kk and the time constant τ\tau.

  10. Ex. 99.10Application

    Derive the formula for kk from two temperature measurements T1T_1 (at t1t_1) and T2T_2 (at t2t_2) with known TambT_{\text{amb}}.

  11. Ex. 99.11ApplicationAnswer key

    T0=100T_0 = 100 °C, Tamb=25T_{\text{amb}} = 25 °C, k=0,05k = 0{,}05 min1^{-1}. Use Euler with h=5h = 5 min to estimate T(15)T(15) and compare with the exact value.

  12. Ex. 99.12Application

    The temperature difference between an object and the environment drops from 80 °C to 40 °C in 10 min. How much additional time until it drops from 40 to 20 °C?

  13. Ex. 99.13Application

    Show that if T0=TambT_0 = T_{\text{amb}}, the solution is constant. Interpret physically.

  14. Ex. 99.14Application

    Milk: T0=72T_0 = 72 °C, Tambcold=0T_{\text{amb}}^{\text{cold}} = 0 °C, k=0,15k = 0{,}15 s1^{-1}. How long to cool to 4 °C?

  15. Ex. 99.15ModelingAnswer key

    Forensic case. Body found at 11:00 PM with T=30T = 30 °C. Tamb=20T_{\text{amb}} = 20 °C, k=0,07k = 0{,}07 h1^{-1}. Estimate the time of death. Discuss the method's uncertainties.

  16. Ex. 99.16Modeling

    Object with constant internal heat source: T=k(TTa)+HT' = -k(T - T_a) + H, where H=Q/(mcp)H = Q/(mc_p). With Ta=22T_a = 22 °C, k=0,05k = 0{,}05 min1^{-1}, H=5H = 5 °C/min. What is the equilibrium temperature?

  17. Ex. 99.17Modeling

    Object warming: measurements T(0)=20T(0) = 20, T(30)=40T(30) = 40, T(60)=55T(60) = 55 °C. Estimate TambT_{\text{amb}} and kk assuming one of the three equations might be noisy.

  18. Ex. 99.18ModelingAnswer key

    Processor with dissipation H=2H = 2 °C/min, Ta=25T_a = 25 °C. To keep T40T \leq 40 °C, what is the minimum kk required in the cooling system?

  19. Ex. 99.19Understanding

    How does the cooling rate T(t)|T'(t)| vary over time for an object with T0>TambT_0 > T_{\text{amb}}?

  20. Ex. 99.20UnderstandingAnswer key

    How does kk depend on the physical properties of the system? What happens to the time constant τ\tau when kk increases?

  21. Ex. 99.21Understanding

    In what situations does Newton's law of cooling cease to be valid?

  22. Ex. 99.22Application

    Two measurements: T(0)=30T(0) = 30 °C, T(1)=28T(1) = 28 °C, Tamb=20T_{\text{amb}} = 20 °C. Determine kk and estimate T(5)T(5).

  23. Ex. 99.23Application

    T(0)=30T(0) = 30 °C, T(1)=28T(1) = 28 °C, Tamb=20T_{\text{amb}} = 20 °C. Determine kk and calculate T(5)T(5).

  24. Ex. 99.24Modeling

    Server: P=2000P = 2000 W, hA=200hA = 200 W/K, Ta=24T_a = 24 °C. What is the equilibrium temperature? What is needed to keep it below 27 °C?

  25. Ex. 99.25ModelingAnswer key

    Ta(t)=20+8cos(πt/12)T_a(t) = 20 + 8\cos(\pi t/12) °C (daily variation with 24 h period). Write the formal solution of T=k(TTa(t))T' = -k(T - T_a(t)) and discuss how the oscillation amplitude of TT compares to that of TaT_a.

  26. Ex. 99.26Proof

    Show that the IVP T=k(TTa)T' = -k(T - T_a), T(0)=T0T(0) = T_0 has a unique solution for all t0t \geq 0.

  27. Ex. 99.27Proof

    Verify by direct substitution that T(t)=Tamb+(T0Tamb)ektT(t) = T_{\text{amb}} + (T_0 - T_{\text{amb}})e^{-kt} satisfies the ODE and the initial condition.

  28. Ex. 99.28ChallengeAnswer key

    Mutual cooling. Two objects exchange heat with each other: T1=k1(T1T2)T_1' = -k_1(T_1-T_2), T2=k2(T2T1)T_2' = -k_2(T_2-T_1). T1(0)=100T_1(0) = 100 °C, T2(0)=20T_2(0) = 20 °C. Find the equilibrium temperature and the rate of approach.

  29. Ex. 99.29Application

    Steel part: T0=850T_0 = 850 °C, Tamb=30T_{\text{amb}} = 30 °C, k=0,02k = 0{,}02 min1^{-1}. How long to cool to 200 °C?

  30. Ex. 99.30Modeling

    Compare Newton's law of cooling with radioactive decay. What are the mathematical similarities? What is the difference in equilibrium?

Sources

  • Lebl, Jiří. Notes on Diffy Qs: Differential Equations for Engineers. Version 6.4. CC-BY-SA. jirka.org/diffyqs — §1.6: Newton's law of cooling as a 1st-order autonomous ODE.
  • OpenStax. Calculus Volume 2. CC-BY-NC-SA. openstax.org/details/books/calculus-volume-2 — §4.4: applications of separable ODEs to Newtonian cooling and forensic estimates.
  • Trench, William F. Elementary Differential Equations with Boundary Value Problems. open. digitalcommons.trinity.edu/mono/9 — §4.2: temperature models with industrial context, exercises with numerical data.

Updated on 2025-05-14 · Author(s): Clube da Matemática

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