Question

Calc @solomonzelman @michele_laino

Answer 1

I need to use Newton's Law of Cooling

Answer 2

@SolomonZelman @zepdrix @Michele_Laino

Answer 3

Room temperature is 70 degrees?
I forget :o

Answer 4

I guess. I think it can be solved without that tho?

Answer 5

If I let current time in minutes, and the temperature that
is exact the room temperature in C to be the origin, then,

\(\color{#000000}{ \displaystyle y(0~{\rm min})=60~{\rm C^\circ} }\)
\(\color{#000000}{ \displaystyle y(-20~{\rm min})=70~{\rm C^\circ} }\)

Time can't be negative in real world, but, this is how I denoted
it, and that shouldn't be such a problem. After all, in physics we
also sometimes call downward the positive direction. Anyway ...

If I could assume that the temperature is linearly changing, then
\(\color{#000000}{ \displaystyle m=\frac{0-20}{60-70}=-2 ~{\rm C^\circ}/{\rm min } }\)

and then I simply have a function to models temperature above room tempterature (hope this doesn't sound too confusing)
\(\color{#000000}{ \displaystyle y=60-2x }\)

Answer 6

It might sound like a problem from a particular appliaction that has its own setup tho'. Like mixing DE problems have (input rate)-(output rate) .... So, I don't really know for sure what kind of change we are talking about here. Linear? then, as I said ... Exponential? (like in population growth...) then it is different.

Answer 7

Sorry, I know almost nothing about chemistry:) My worst subject.

Answer 8

Yeah, I'm pretty sure it is supposed to be exponential

Answer 9

Have you ever heard of Newton's Law of Cooling?

Answer 10

No, actually:) I only know Newton's Laws that relate to the subject, Physics.

Answer 11

So, if you sure it is exponential, then I am almost positive that it is modeled by some generic exponential function in a form of:
\(\color{#000000}{ \displaystyle f(x)=a\cdot e^{kx} }\)

(Which is a solution to a typical popul. prob. \(\color{#000000}{ \displaystyle \frac{dp}{dt}=kt }\))

Answer 12

The formula in my book is
\(\Large T-T_s=(t_0-t_s)e^{-kt}\)

T = temperature of the object at time t
Ts = temperature of surrounding
T0 = temperature at time 0

Answer 13

ps those ts are supposed to be uppercase

Answer 14

Well, you have the given info...
so if your room temperature is say \(\varphi\) degrees (I like greek letters),
(but, you shall determine (find online, or something ...) the room
temperature. I am just to making point regarding the math of it ...)

Lowercase t, time Uppercase T, Temperature.
\(\color{#000000}{ \displaystyle t=0~~~~~~~~~~~~~~{\rm T}=\varphi+60 }\)
\(\color{#000000}{ \displaystyle t=-20 ~~~~~~~~~{\rm T}=\varphi+70 }\)

I am using the current time as t=0,
and \(\varphi\) is the room temperature (whatever you find it to be online in your book, or wherever....), so \(\varphi+60\) is just 60 degrees above room Temperature.

Yeah .... I might have written it more complexly, but just didn't want to seem like I am actually solving the problem with known room Temperature, when I in fact don't know it.

Can you re-write the formula tho? You said t's are all capitalized, or what?

Answer 15

That's the formula?
\(\color{#000000}{ \displaystyle T-T_S=(T_0-T_S)e^{-kt} }\)

Answer 16

yes

Answer 17

so you don't technically need the value of room temp to solve?

Answer 18

Oh, then you plug it in:
\(\color{#000000}{ \displaystyle T-T_S=(T_0-T_S)e^{-kt} }\)
\(\color{#000000}{ \displaystyle T(t)=T_S+(T_0-T_S)e^{-kt} }\)

I'm just writing it in a way that we know that Temp. is a function of time.
Then, we can plug in the information we know:
\(\color{#000000}{ \displaystyle T(t)=\varphi +(60+\varphi-\varphi)e^{-kt} }\)

and no, I didn't say that, I am using \(\varphi\) because \(\underline{~I~}\) don't know what the room temperature is.

Answer 19

And, next I know that
\(\color{#000000}{ \displaystyle T(-20)=\varphi+70 }\)

(In other words: 20 minutes ago, it was 70 degrees above room tempterature)

And you can use that info to find k.

Answer 20

So, having looking up the room temperature (which I wrote as \(\varphi\)), you can redo what I did but without writing \(\varphi\) for room temperature.

You will be able to find k, logarithmically (just trig, or algebra 2 ...)

Answer 21

So after this you have a function for temperature in terms of time. And now, you don't have any unknown values (you can approximate \(e\) however you like) ... so you can answer your questions a-c from here.

Answer 22

Just don't want to reword it myself, cuz I hate algebra when it's not necessary for me. And why, if you can do it just as well? :)

Answer 23

if you have questions, I will ty to answer them timely. good luck

Answer 24

Thank you so much for your help! sorry for the late reply. openstudy lag

Answer 25

Yeah, it really does lag ... I checked it is not an individual computer issue.

Answer 26

:)

Answer 27

one question. why do you have that weird symbol minus the weird symbol?

Answer 28

\(\color{#000000}{ \displaystyle T(t)=\varphi +(60+\color{red}{\varphi-\varphi})e^{-kt} }\)