Cooling of coffee

Example — Cooling of coffee (Newton's law)

A cup of coffee at temperature $T$ cools in an environment at $T_{\text{amb}}$: $\frac{dT}{dt} = -k\bigl(T - T_{\text{amb}}\bigr), \qquad k>0.$ We separate: $\dfrac{dT}{T-T_{\text{amb}}} = -k\,dt$. Integrating: $\ln|T-T_{\text{amb}}| = -kt+C$, hence: $\boxed{T(t) = T_{\text{amb}} + (T_0 - T_{\text{amb}})\,e^{-kt}}$

The temperature decreases exponentially towards $T_{\text{amb}}$, at a rate proportional to the current difference between the temperature of the body and that of the environment.

Links

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