Newton's law of cooling

About this lesson

Calculus AB (AP-aligned) › Differential equations

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Newton's law of cooling

Warm-up

Three quick questions from earlier lessons.

Discover

A cup at temperature bb °C cools in a 20 °C room. The violet line is the room temperature. Move bb and the cooling constant kk.

Newton's law of cooling: T = 20 + (b − 20)e^(−kt). Whatever the start, the temperature approaches the room's 20 °C, faster for larger k.

Which temperature does the cup approach?

  1. 20 °C
  2. 0 °C

Key idea

The difference with the surroundings decays exponentially.

Newton's law of cooling: dTdt=−k(T−A)\dfrac{dT}{dt} = -k(T - A), with AA the surrounding temperature and k>0k > 0. Solution: T=A+(T0−A)e−ktT = A + (T_0 - A)e^{-kt}.

Worked example

Tea at 90 °C in a 20 °C room, k=0.1k = 0.1 per minute.

T−20=70e−0.1tT - 20 = 70e^{-0.1t}
u = T − 20 satisfies u' = −0.1u.
T=20+70e−0.1tT = 20 + 70e^{-0.1t}
So:
It starts cooling at 7 °C per minute
At t = 0 the rate is −0.1 × 70.

Your turn, guided

Your turn: soup at 80 °C in a 25 °C room, k=0.2k = 0.2. The rate at t=0t = 0?

−11-11 °C/min
−0.2 × (80 − 25).

Practice ★

  1. Soup at 60 °C is left in a 17 °C kitchen and follows Newton's law of cooling. Which temperature does it approach in the long run?
  1. dTdt=−410(T−25)\dfrac{dT}{dt} = -\tfrac{ 4 }{10}(T - 25) and T(0)=74T(0) = 74. How fast is the temperature changing at t=0t = 0, in °C/min?
  2. dTdt=−410(T−22)\dfrac{dT}{dt} = -\tfrac{ 4 }{10}(T - 22) and T(0)=92T(0) = 92. How fast is the temperature changing at t=0t = 0, in °C/min?

Watch out!

Practice ★★

  1. A drink at 65 °C cools in a 18 °C room: dTdt=−310(T−18)\dfrac{dT}{dt} = -\tfrac{ 3 }{10}(T - 18). Find T(t)T(t).
  2. A drink at 89 °C cools in a 16 °C room: dTdt=−210(T−16)\dfrac{dT}{dt} = -\tfrac{ 2 }{10}(T - 16). Find T(t)T(t).

Exit ticket

Three questions to check you have got it.

  1. A drink at 65 °C cools in a 18 °C room: dTdt=−310(T−18)\dfrac{dT}{dt} = -\tfrac{ 3 }{10}(T - 18). Find T(t)T(t).
  2. A drink at 89 °C cools in a 16 °C room: dTdt=−210(T−16)\dfrac{dT}{dt} = -\tfrac{ 2 }{10}(T - 16). Find T(t)T(t).
  3. A drink at 79 °C cools in a 15 °C room: dTdt=−210(T−15)\dfrac{dT}{dt} = -\tfrac{ 2 }{10}(T - 15). Find T(t)T(t).

Summary card

Newton's law of cooling

  • T′=−k(T−A)T' = -k(T - A).
  • T=A+(T0−A)e−ktT = A + (T_0 - A)e^{-kt}, approaching AA.

Watch out: it is the difference T−AT - A that decays.

End of the lesson

Back to the chapter

Answers: Newton's law of cooling

A

  1. 1717

B

  1. −985- \frac{98}{5}
  2. −28-28

C

  1. 18+47e−3t1018 + 47 e^{- \frac{3 t}{10}}
  2. 16+73e−t516 + 73 e^{- \frac{t}{5}}

D

  1. 18+47e−3t1018 + 47 e^{- \frac{3 t}{10}}
  2. 16+73e−t516 + 73 e^{- \frac{t}{5}}
  3. 15+64e−t515 + 64 e^{- \frac{t}{5}}

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