How long will it take for the object to cool to 20 K?

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: The rate that an object cools is directly proportional to the difference between its temperature (in Kelvins) at that time and the surrounding temperature (in Kelvins). If an object is initially at 35 K, and the surrounding temperature remains constant at 10 K, it takes 5 minutes for the object to cool to 25 K. How long will it take for the object to cool to 20 K?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by chisigma, I like Serena, lfdahl, magneto, MarkFL, and Sudharaka. You can find I like Serena's solution below.

[sp]The problem corresponds to the differential equation
$$\dot T = -k(T - T_e)$$
where $T_e$ is the exterior temperature.

The solution is:
$$T = (T_0 - T_e) e^{-t/\tau} + T_e$$
where $T_0$ is the initial temperature and $\tau$ the so called characteristic time.

Filling in the numbers we find that:
$$T = (35 - 10) e^{-t/\tau} + 10 = 25 e^{-t/\tau} + 10$$

Applied to $T(5\text{ min}) = 25$ we find that:
$$\tau = \frac{5\text{ min}}{\ln(5/3)}$$

Solving for $T=20$, we get that:
$$25 e^{-t/\tau} + 10 = 20$$
$$t = 5\text{ min} \cdot \frac {\ln(5/2)}{\ln(5/3)} \approx 9 \text{ min}$$

$\blacksquare$[/sp]