DE: Temperature and Coffee

**calc101** · June 26th 2009, 06:54 AM

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:

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

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

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

$<br /> t = - ln(\frac {78 - 20} {95})<br />$

Therefore, the Coffee's temperature will be decreasing at 1 degree per minute after: 0.4934 minutes.

**skeeter** · June 26th 2009, 10:06 AM

Newton's Law of cooling ...

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

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

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

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

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

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

$T-20 = Ae^{kt}$

$T = 20 + Ae^{kt}$

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

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

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

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

$58 = 75e^{kt}$

$t = \frac{1}{k} \cdot \ln\left(\frac{58}{75}\right) \approx 15$ minutes

**calc101** · June 26th 2009, 12:44 PM

I see my mistake!

Thanks a lot Skeeter,

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