DE: Temperature and Coffee

• Jun 26th 2009, 06:54 AM
calc101
DE: Temperature and Coffee
Problem:

A freshly brewed cup of coffee has temperature 95°C in a 20°C room. When its temperature is 78 °C, it is cooling at a rate of 1°C per minute. When does this occur? Approximate to three decimal places.

Here is my work, please verify if it is correct:

$\displaystyle T(t) = y0 * e^{kt} + Ts$

$\displaystyle 78 = 95*e^{kt}+20$

$\displaystyle 95e^{-t} = 78 - 20$

$\displaystyle t = - ln(\frac {78 - 20} {95})$

Therefore, the Coffee's temperature will be decreasing at 1 degree per minute after: 0.4934 minutes.
• Jun 26th 2009, 10:06 AM
skeeter
Newton's Law of cooling ...

$\displaystyle \frac{dT}{dt} = k(T - T_a)$

$\displaystyle \frac{dT}{dt} = k(T - 20)$

when $\displaystyle T = 78$ , $\displaystyle \frac{dT}{dt} = -1$

$\displaystyle k = -\frac{1}{58}$

$\displaystyle \frac{dT}{T-20} = k \, dt$

$\displaystyle \ln|T-20| = kt + C$

$\displaystyle T-20 = Ae^{kt}$

$\displaystyle T = 20 + Ae^{kt}$

at $\displaystyle t = 0$, $\displaystyle T = 95$ ...

$\displaystyle 95 = 20 + A$ ... $\displaystyle A = 75$

$\displaystyle T = 20 + 75e^{kt}$

$\displaystyle 78 = 20 + 75e^{kt}$

$\displaystyle 58 = 75e^{kt}$

$\displaystyle t = \frac{1}{k} \cdot \ln\left(\frac{58}{75}\right) \approx 15$ minutes
• Jun 26th 2009, 12:44 PM
calc101
I see my mistake!

Thanks a lot Skeeter,