# Newton's Law of Cooling differential equation

1. Sep 5, 2008

### 2RIP

Problem
From Newton's Law of Cooling, we can use the differential equation

dT/dt= -k(T-Ts)

where Ts is the surrounding temperature, k is a positive constant, and T is the temperature.

Let $$\tau$$ be the time at which the initial temperature difference T0-Ts has been reduced by half. Find the relation between k and $$\tau$$.

Work

I have found the temperature equation for the object to be

T(t) = Ts + (T0-Ts)e-kt

If the initial temperature difference is reduced by half, i tried

T(t) = Ts + 0.5(T0-Ts)e-kt

but couldn't solve it any further. Could someone please shed some light on what i should do next?

2. Sep 6, 2008

### jostpuur

This seems to be right, but here:

you made some mistake with the parameter t. If we set notation

$$\Delta T(t) = T(t) - T_s,$$

then the equation you want to solve is

$$\Delta T(\tau) = \frac{1}{2}\Delta T(0).$$

3. Sep 7, 2008

### 2RIP

Oh thanks for replying, i was able to get the correct answer.

k $$\tau$$ = ln2

However, I'm still a bit unclear of why T($$\tau$$)-Ts and T0-Ts canceled out during calculation. Or do T($$\tau$$) and T0 equal? Also why isn't the 0.5 placed like

T($$\tau$$) = Ts + 0.5(T0-Ts)e-k$$\tau$$, but instead

0.5[T($$\tau$$)-Ts] = (T0-Ts)e-k$$\tau$$

Thanks again!

Last edited: Sep 7, 2008
4. Sep 7, 2008

### jostpuur

The use of latex like this doesn't seem to be working very well. At least my browser makes that look like $$k^{\tau}=\textrm{ln}(2)$$, although you are most certainly meaning $$k\tau = \textrm{ln}(2)$$.

It looks like you are getting the correct result, because you know what it is, and you are arriving at it "by force". Just try to chop the problem into smaller pieces, so that you can control it. We want this:

$$\Delta T(\tau) = \frac{1}{2} \Delta T(0)\quad\quad\quad\quad (1)$$

where the temperature difference at each instant is given by

$$\Delta T(t) = T(t) - T_s = (T_0-T_s) e^{-k t}.$$

So onto the left side of the equation (1) we get

$$(T_0 - T_s)e^{-k\tau}$$

and on the right side

$$\frac{1}{2} (T_0 - T_s)e^{-k\cdot 0}.$$

See how things go then?

5. Sep 7, 2008

### 2RIP

Oh yeah i see it now, thanks a lot jostpuur. As for the latex inputs, I think I will go look for some tutorials on this forum. :rofl: