Question

The temperature of a cup of hot tea varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the tea, A is the room temperature, and k is a positive constant. If the water cools from 100°C to 80°C in 1 minute at a room temperature of 60°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.

Answer 1

https://gyazo.com/8aa6a713d1c912933a9dcf9581c5483f

Answer 2

@jim_thompson5910 you taught me this a while back. I used your method and got 58 it would be nice if you can check it

Answer 3

work with the temperature differences
find the exponential model that goes through the points \((0,40)\) and \((1,20)\)

Answer 4

@satellite73 could you finish helping me with my question?

Answer 5

initially the difference is 40, after one minute it is 20

Answer 6

@satellite73 I am not sure what you're trying to get at
I followed http://openstudy.com/users/ephemera#/updates/56468ffde4b0c3aa4a128f95

Answer 7

ok I'm looking it over now

Answer 8

@Ephemera can you post your work so far?

Answer 9

@Ephemera is your answer 58 ?
the room temperature is 60, how can the temperature of tea go below room temperature ?

Answer 10

Site was rally glitchy for me. And I know what I was doing wrong @ganeshie8 I used the same exact formula when in that question they ask for time, here they ask for temperature. According to your reasoning (very good reasoning might I add) the answer can't be 58. Please allow me a couple minutes to try and do it properly.

Answer 11

Ok so this is what I have:
\[\left| T-A \right| = e^c e^-kt\]
T = 100
A= 80
t= 1
e^c = 60?

Answer 12

\[\left| 100-80 \right| = 60e ^{-k}\]

Answer 13

is it good so far?

Answer 14

@jim_thompson5910 @ganeshie8

Answer 15

I'd solve for k then use k to find T?

Answer 16

dt/(T-A) = -kdt
ln(|T-A|) = -kt + C
|T - A| = e^(-kt + C)
|T - A| = e^C*e^(-kt)
|T - 60| = e^C*e^(-kt)
|100 - 60| = e^C*e^(-k*0)
40 = e^C
e^C = 40

I'm not sure how you got e^C = 60

Answer 17

hmm

Answer 18

Remember, \(T\) is a function of time \(t\). Your function for temperature at any given time should be :
\[T(t) = 60 + Be^{-kt}\]

Answer 19

What does B represent?

Answer 20

B is a constant

Answer 21

B = e^C

Answer 22

plugin (0, 100) in above equation and you may solve B

Answer 23

Same as e^c, I see it now

Answer 24

So we need to solve for k then T now, correct?

Answer 25

T is not a constant

Answer 26

T is the temperature at any given time

Answer 27

Are they not asking for T(4)?

Answer 28

Yes, they are

Answer 29

And I can't solve for that without k correct?

Answer 30

Yes

Answer 31

That is what I meant.

Answer 32

find the constants first using the given initial conditions

Answer 33

I would use (1,80) to solve for k?

Answer 34

Yes, try finding B first though

Answer 35

I thought be was equal to 40

Answer 36

B*

Answer 37

Or would it be a different scenario for 80?

Answer 38

yes, B = 40

Answer 39

B is a constant, it doesn't change

Answer 40

\[k= -\ln(1/2)\]

Answer 41

or equivalently, k = ln(2)

Answer 42

Yes using log properties.

Answer 43

I'm getting T(4) = 700

Answer 44

way too high

Answer 45

exactly

Answer 46

\[T(4) = 60 + 40e^{4\ln2}\]

Answer 47

you should find that
\[\Large T(t) = 60+40e^{-\ln(2)*t}\]

Answer 48

you lost a negative in your exponent

Answer 49

why would there be a negative?

Answer 50

-ln(1/2) ------> ln(1/2)^-1

Answer 51

\[\Large T(t) = 60+40e^{-kt}\]
\[\Large T(t) = 60+40e^{-\ln(2)t}\]

Answer 52

Which is ln 2

Answer 53

Ohhhhh

Answer 54

The negative in the formula, got you

Answer 55

62.5 or 63

Answer 56

yep

Answer 57

thanks for both your help

Answer 58

np