Newton's Law of Cooling question

Quote:

Originally Posted by Kakariki

Hey guys, I am in a calculus course, and this is the final question of the lesson on related rates.

question
A thermometer has been inside a roast cooking in the oven. Just before the thermometer was removed from the roast, it read 180C. The thermometer is removed and left on the kitchen counter to cool just before putting it in the dishwasher. After one second, it has cooled 8C. After two seconds, it has cooled 26C more. What is the room's temperature?

Solution
Okay, they give me the equation for Newton's law of cooling: $T - S = Ce^{kt}$
Where: T represents the final temperature of the object, C represents the initial temperature of the object, S is the temperature of the surroundings, k is the constant of proportionality, and t is time.

You mean that T is the temperature at time t, not necessarily the "final" temperature. But C cannot be "the initial temperature of the object" because, at time t= 0, we have T(0)- S= C. T(0) is the "initial temperature of the object" because "initial" means t= 0.

C is the difference between the the initial temperature and room temperature. C is NOT 180.

Quote:

So I basically need to find S. My attempt is basically solving a system of equations, and I really doubt this is how to do it because my answers have been completely wrong.

I need ALOT of help in this problem. All I have so far is filling in the values for everything but S and k in the equations, and I do not have the necessary math skills to solve the system of equations to find what k and S are.

$S = -180e^k + 172$
Plug this into the other equation:
$-26 + 180e^k = 180e^{2k}$

Your first equation is wrong because, as I said, C is NOT "the initial temperature".

At t= 0, T= 180 so 180- S= C.

At t= 1, T= 180- 8= 172 so $172- S= Ce^{k}$ .

At t= 2, T= 172- 26= 146 so $146- S= Ce^{2k}$ .

Those are the three equations you need to solve for S, C, and k.

One thing you can do, of course, is to replace "C" in the last two equations by 180- S:

$172- S= (180- S)e^{k}$
$146- S= (180- S)e^{2k}$

Seeing "k" and "2k" in the exponent, I would try squaring both sides of the first equation:
$(172- S)^2= (180- S)^2e^{2k}$
$\frac{(172- S)^2}{180- S)= (180- S)e^{2k}$

And since that and $146- S$ are equal to $(180- S)e^{2k}$ you can set them equal to get a single equation for S.

Newton's Law of Cooling question - need major help