Exponential Decay (1 Viewer)

Lukybear · Dec 24, 2009

A heated body is cooling and the excess of its temperature, x, above that of its surroudnings is x=Ae^(-kt) where t is time in minutes

a)At time t=0, x = 80 find A
b)If the temperature of the surroudnings is 20, and the body cools to 70 in 10min, find

i) Temperature after 20min
ii) Time taken to reach 60

I couldnt get this... and yet it seems so easy

Ive got A=80
and for bi) by using calculator ive got 31.25

my k was something like 0.04.... by calculator ofcoruse

Did anyone see my mistake or is the book wrong?

Cazic · Dec 24, 2009

The body cools to 70 in 10 mins (with a surrounding temp of 20), so:

$\begin{align*}&\hskip 1.5em x(10) = 70 - 20 = 50 = 80e^{-10k} \\ &\Rightarrow k = \frac{\ln(5/8)}{-10} \\ &\Rightarrow x(t) = 80e^{\frac{\ln(5/8)t}{10}} = 80\left(\frac{5}{8}\right)^{\frac{t}{10}}.\end{align*}$

Now, the temperature of the body after 20 minutes is the surrounding temp + excess temp, that is:

$20 + x(20) = 20 + 80\left(\frac{5}{8}\right)^{2} = 51.25.$

Finally, the body is 60 degrees when x(t) = 40, that is,

$\begin{align*}x(t) = 40 &\Rightarrow 80\left(\frac{5}{8}\right)^{\frac{t}{10}} = 40 \\ &\Rightarrow t = 10 \log_{\frac{5}{8}}\left(\frac{1}{2}\right) = 10 \frac{\ln(1/2)}{\ln(5/8)} \approx 14.75.\end{align*}$

Or something like that.

Exponential Decay (1 Viewer)

Lukybear

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Cazic

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