Question

Can anyone help me with setting up a Newtons law of cooling problem?

Answer 1

what is the question?

Answer 2

My textbook says it id dT/dt=k(T-Tm) but my professor implied it's dT/dt=k(Tm-T). What is it?

Answer 3

depends if item is heating or cooling, and if k is positive or negative

Answer 4

if the item is cooling, you want the right side to be negative and otherwise positive

Answer 5

In the problem in the book a cake was cooling and in the class it was a cup of coffee cooling.

Answer 6

I'll put up the problem, give me a sec. Thanks

Answer 7

then it should be T' = -k(cake - room)

Answer 8

When a cake is removed from an oven, its temperature is measured at 300F. Three minutes later its temperature is 200F. How long will the cake take to cool off to a temperature of 70F?

Answer 9

That's right, the equation the give is T'=k(T-70). I'll put up the other problem

Answer 10

thats assuming the room is 70 degrees?

Answer 11

The cup of coffee originally at 95C and cools to 80C in five minutes while siting in a room of temperature of 21C. Find a DE...

Answer 12

Yeah the room is 70 degrees, I left out a word

Answer 13

He said the equation is DT/dt=k(21-T)?

Answer 14

T = 70 + (300 - 70)e^(kt)

Answer 15

You don't have to solve the equation, we just have to set it up

Answer 16

that works too, just means the difference between the coffee and ambient will be negative, so k will be positive

Answer 17

So what is the total equation by the coffee?

Answer 18

-k(cake - room) == k(room - cake)

Answer 19

need to find k with initial values

Answer 20

don't worry about the solution, we're just supposed to set up the problem

Answer 21

200 = 70 + (230)e^(k * 3)

Answer 22

How is the coffee equation DT/dt=k(21-T), shouldn't it be (T-21)

Answer 23

that was for the cake, btw, not coffee

Answer 24

How does the coffee work?

Answer 25

it works either way so long as your constant k's sign reflects whether it is cooling or heating

Answer 26

I'm missing something, here by DT/dt=k(21-T) there is no negative so shouldn't it be the opposite

Answer 27

So you're saying that the signs of the k's would be opposite?

Answer 28

comparing the cake to the coffee

Answer 29

The cup of coffee originally at 95C and cools to 80C in five minutes while siting in a room of temperature of 21C. Find a DE...

T = 21 + (95 - 21)e^(-kt)
so
80 = 21 + 74e^(-k*(5))

Answer 30

use that to solve for k, then stick that k back in the first one

Answer 31

So basically, when it comes down to it it can be either k(T-Tm) or k(Tm-t) and the equation will come out the same. Is that right?

Answer 32

so whats Tm and whats t?

Answer 33

Tm is room temperature and T is material temperature

Answer 34

sorry, it should be capitalized

Answer 35

-k(cake - room) == k(room - cake)

-k(T - Tm) == k(Tm - T)

Answer 36

But since the k is unknow right now, can I just switch them and it will be automatically reflected in the k

Answer 37

does that make sense?

Answer 38

because (cake - room) will be positive if the cake is hotter than than the room, but the cake is cooling so dT/dt will have to be negative, therefore k needs to be negative

Answer 39

But do I have to put the negative there or will that happen automatically when I solve for k?

Answer 40

no, pay attention to it, make sure it is the right sign in the T' eq

Answer 41

if I were you i would just set up the T' eq so k is always neg, "-k"

Answer 42

But the professor just wrote the equation as dT/dt=k(21-T).
Is he mistaken?

Answer 43

he didn't put in a negative

Answer 44

idk, i would ask him if its important

Answer 45

I wish I could but I have a test today heh heh
In your opinion though, is it wrong?

Answer 46

I would do it the way your professor gave it to you. as long as you understand how the derivative works (positive it heating, neg if cooling), you should be fine

Answer 47

Alright, thanks I really appreciate it!

Answer 48

no prob, good luck on your test