Problem — Cooling of a cup of tea

Problem

A cup of tea at $90°$C is in a room at $20°$C; after 5 minutes the tea is at $70°$C. Find $k$ and calculate the time to reach $30°$C.

Solution

By Newton’s law of cooling $T(t) = T_{\text{amb}} + (T_0-T_{\text{amb}})e^{-kt} = 20 + 70\,e^{-kt}$.

Calculation of $k$. From $T(5)=70$: $70 = 20 + 70\,e^{-5k}$, that is $e^{-5k} = \dfrac{50}{70}=\dfrac{5}{7}$. Hence: $k = \frac{1}{5}\ln\frac{7}{5} \approx 0{,}0673\ \text{min}^{-1}.$

Time to reach $30°$C. From $30 = 20 + 70\,e^{-kt}$: $e^{-kt} = \dfrac{10}{70}=\dfrac{1}{7}$, hence: $t = \frac{\ln 7}{k} = \frac{\ln 7}{0{,}0673} \approx \boxed{28{,}9\ \text{min}}$

Links

Topics: Differential equations
Concepts: Decay · Newton’s law of cooling · Separable variables
Functions: Exponential function
Skills: Calculating · Modelling
People: Isaac Newton
Exercise type: Modelling problem