# Newtons law of cooling

## Homework Statement

Newton's law of cooling says that the rate at which the emperature of any object changes is proportional to the temperature difference between the object and its surroundings. So if the temperature of the object is T, and its surroundings are at a constant temperature $$\hat{T}$$

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

a) let $$\theta = T - \hat{T}$$, Show that $$\frac{d\theta}{dt} = -k\theta$$ and hence write down the general solution for $$\theta(t)$$

b) given $$\theta(0) = 64$$ and $$\theta(2) = 36$$, find $$\theta(3)$$

## The Attempt at a Solution

a)

$$\frac{d\theta}{dt} = -k\theta$$

$$\int \frac{d\theta}{dt} = \int -k\theta$$

re arranging
$$\int \frac{d\theta}{\theta} = \int -k dt$$

and integrating
$$ln(\theta) = -k t + C$$
(I think I add a + C?)

exponentiating both sides
$$eln(\theta) = e^{-kt +c} = Ce^{-kt}$$
though I'm not sure why the C goes out to the front, or why it is being multiplied by C

simplifying gives the solution
$$\theta(t) = Ce^{-kt}$$

b)

$$\theta(0) = Ce^{-k * 0} = 64$$

since $$e^{-k * 0} = e^0 = 1$$
that makes
$$C*1 = 64$$
so C = 64

now
$$\theta(2) = 36$$
so

$$64*e^{-k * 2} = 36$$

$$e^{-k * 2} = \frac{36}{64}$$

taking the logarithim of each side

$$-k*2 = ln(\frac{36}{64}$$

solving for k, I get, K = 0.28768

then going to $$\theta(3)$$

putting it into the equation, then a calculator

64*e^{-0.28768*3} = 27

I think this is right, since it is a real number / integer, and it sort of fits the pattern

-

if someone could check, i'd be very happy, thank you :P

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Mentallic
Homework Helper
Firstly, you haven't answered the very first question.

a) let $$\theta = T - \hat{T}$$, Show that $$\frac{d\theta}{dt} = -k\theta$$
Don't worry, when I was studying this, I did exactly the same thing and assumed I already had answered it. It's quite an arbitrary idea so you may need a little more practise here, because it's asked quite often on tests.

So you have the equations:

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

$$\theta=T-\hat{T}$$

And you need to show that

$$\frac{d\theta}{dt} = -k\theta$$

Use the fact that $$\frac{d\theta}{dt}=\frac{d\theta}{dT}.\frac{dT}{dt}$$

exponentiating both sides
$$eln(\theta) = e^{-kt +c} = Ce^{-kt}$$
though I'm not sure why the C goes out to the front, or why it is being multiplied by C
By a rule of exponentiation, $$a^{b+c}=a^ba^c$$

so $$e^{-kt+c}=e^{-kt}e^c$$ but since c is just some constant we don't know yet, ec is also just some constant, so we simplify things and just give it another constant C (big c, don't use the same constant because $c\neq e^c$).

Everything else is right, but if you aren't strapped for time in a test situation, then you may want to try a slightly more elegant approach to giving the final answer which will be an exact solution rather than relying on your calculator, for example, rather than:

taking the logarithim of each side

$$-k*2 = ln(\frac{36}{64}$$

solving for k, I get, K = 0.28768

then going to $$\theta(3)$$

putting it into the equation, then a calculator

64*e^{-0.28768*3} = 27
You could do this,

$$-2k=ln\left(\frac{36}{64}\right)=ln\left(\frac{9}{16}\right)$$

$$k=\frac{-1}{2}ln\left(\frac{9}{16}\right)$$

$$=ln\left(\left(\frac{9}{16}\right)^{-1/2}\right)$$

$$=ln\left(\left(\frac{16}{9}\right)^{1/2}\right)$$

$$k=ln\left(\frac{4}{3}\right)$$

then

$$\theta(3)=64e^{-3ln\left(\frac{4}{3}\right)}$$

$$=64\left(e^{ln4-ln3}\right)^{-3}$$

by using a combination of the rules $$ln\frac{a}{b}=lna-lnb$$ and $$e^{ab}=(e^a)^b$$

$$=64\left(\frac{e^{ln4}}{e^{ln3}}\right)^{-3}$$

$$=64\left(\frac{3}{4}\right)^3$$

$$=64\left(\frac{27}{64}\right)$$

$$\theta(3)=27$$

There you have it, a confirmation that 27 is indeed the correct answer and your calculator isn't rounding off or anything.
Doing this in practise you'll of course be able to skip a lot of steps when you get more comfortable at doing it good luck!