Newton's law of cooling: water to 40°C

Let T be the temperature of the water at time t (minutes) and let the room temperature be 30°C. Newton's law gives

dT/dt = −k(T − 30), k > 0

Solve

dT/(T − 30) = −k dt → ln(T − 30) = −kt + ln C

T − 30 = C e^(−kt)

At t = 0, T = 100: 70 = C. So T = 30 + 70 e^(−kt).

Find k

At t = 10, T = 80: 50 = 70 e^(−10k), so e^(−10k) = 5/7 and

k = (1/10) ln(7/5) = 0.03365 per minute.

Time to reach 40°C

40 = 30 + 70 e^(−kt) → e^(−kt) = 1/7

kt = ln 7 → t = ln 7/k = 10 ln 7/ln(7/5) = 10(1.9459)/(0.33647)

t ≈ 57.8 minutes (about 58 minutes after the start, or 47.8 minutes after the 80°C reading).