Question

@tkhunny If you pour a cup of coffee that is 200 F, and set it on desk in a room that is 68 F, and 10 minutes later it is 145 F, what temperature will it be 15 minutes after you originally poured it?

Answer 1

You will need a model of some sort. Are we using Newton's Law of Cooling or some simpler function for our cooling process?

Answer 2

In my class, we talked about Newton's Law of Cooling. The formula is T = A + Se^(-kt) right?

Answer 3

There you go. Please assign values to the known parameters.

T = ??
A = ??
S = ??
e = ??
k = ??
t = ??

Answer 4

T= Would this be the temperature after 15 minutes?
A= 68 F
S= 200 F
k= constant?
t= 15 minutes

Answer 5

A is the Ambient Temperature. This is 68ºF
S is the Initial Difference in Temperature. This is 200ºF - 68ºF = 132ºF

These are the two important parameters to know with surety.

Our task is to find 'k'. This is normally done with a hint. It had better be given in teh problem statement.

" and 10 minutes later it is 145 F, "

THERE it is!!

Knowing that T(10 min) = 145ºF = 68ºF + 132ºF * e^(-k*(10 min)), solve for k and you'll be nearly done.

Answer 6

Ah! Most of it makes sense, but solving for k is confusing to me. How do I solve for it?

Answer 7

You should have enough logarithms in your background. No?

Answer 8

I'm really not good at math, so no, I'm not good with any of this.

Answer 9

Then you cannot solve this problem. There should have been prerequisites to this course. Those prerequisites would have included simple problem solving and logarithms.

Solve for P

112 = 61 + 97P

Answer 10

P= 51/97 or P=52

You wanted me to solve that right? Or was it an example?

Answer 11

I did intend for you to do that. Hopefully, you meant P = 0.52

Okay, now go solve your problem. It requires EXACTLY the same process.

Answer 12

So, T(15)= 126.9 F?

And yes, I did mean 0.52 whoops.