Differential Equations

Newton’s cooling — decay with an offset

continues from lesson 1 — values defined earlier in the course stay live here

Coffee doesn't cool toward zero, it cools toward the room: T′ = −k(T − T_room). The solution is the same exponential, measured from the ambient line. How long until 90 °C coffee is drinkable (60 °C) in a 20 °C room, with k = 0.05/min?

T_{room} = 20

°C

T_{0} = 90

fresh coffee

k_{c} = 0.05

per minute — depends on the cup

T_{cup}\left(t\right) = 20 + 70 \cdot \mathrm{exp}\left(-0.05 \cdot t\right)

temperature over time

t_{drink} = \frac{\mathrm{log}\left(\frac{70}{40}\right)}{0.05}

11.1923

solve T(t) = 60 for t: ≈ 11 min

✓ pass

\mathrm{abs}\left(\mathrm{T_{cup}}\left(t_{drink}\right) - 60\right) < 0.01

at t_drink it is exactly 60 °C