Hot Cup of Joe

Using Newton's law of cooling we can express the temperature $\theta$ at any time $t$ in terms of the initial temperatue $\theta_i$ , ambient temperature $\theta_a$ as $\theta =\theta_a+(\theta_i-\theta_a)e^{-αt}$ , where $α$ is a constant. Here $\theta=60, \theta_i=89, \theta_a=22$ . Also, at $t=1, \theta=85$ . Therefore $α=\ln {\dfrac{67}{63}}$ . Hence we get the required time as $t=\dfrac{\ln {\dfrac{67}{38}}}{\ln {\dfrac{67}{63}}}=\boxed {9.212...}$

Performing an energy rate balance over the boundary of the coffee mug/ environment ( neglecting heat transfer by radiation ):

$\displaystyle \dot{E_{st}} = -\dot{E_{out}}$

$\displaystyle m c_p \frac{dT}{dt} = -h A \left(T - T_{\infty} \right)$

$m$ is mass of coffee

$c_p$ is the specific heat of coffee

$T$ is the instantaneous temperature of the coffee

$T_{\infty}$ ambient temperature of the surroundings

$h$ convection coefficient

$A$ boundary surface area

$t$ is time

The resulting differential equation is linear, first order, non-homogenous. However, making the follwing substitution transforms it into type linear, first order, homogenous:

$\theta = T - T_{\infty}$ it follows that $\frac{d \theta}{dT} = 1$ or $d\theta = dT$

This is readily solved by separation of variables and integration.

Let $\displaystyle K = \frac{h A}{m c_p}$

$\displaystyle \int_{\theta_o}^{\theta_f} \frac{d \theta}{\theta} = -K \int_{0}^{t_f} dt$

$\displaystyle \text{ln} \left( \frac{ \theta_f }{ \theta_o } \right) = -Kt$

The objective of this step is to empiracally solve for the coefficient $K$ using the given experimental data.

$\theta_o$ and $\theta_f$ are respectively the inital and final temperature diffrences between the coffee and the environment.

$\displaystyle K = -\text{ln}\left( \frac{\text{85-22}}{\text{89-22}} \right) \left[ \frac{\text{1}}{\text{min}}\right]$

The last step is to substitute this result back into the solution and determine the time for cooling to $\text{60 C}$

$\displaystyle t_{60} = \frac { \text{ln} \left( \frac{ \text{60-22} }{ \text{89-22} }\right) } { \text{ln} \left( \frac{ \text{85-22} } { \text{89-22} }\right) } \approx \text{9.21 min}$