Question

You place a cup of 200 degrees F coffee on a table in a room that is 67 degrees F, and 10 minutes later, it is 195 degrees F. Approximately how long will it be before the coffee is 180 degrees F? Use Newton's law of cooling:

T(t)=T[a]+(T[o]-T[a])e^-kt

A. 40 minutes
B. 15 minutes
C. 1 hour
D. 35 minutes

Answer 1

PLEASE HELP !! i cant get this wrong .. i really need to know the process

Answer 2

and answer ...

Answer 3

I had a similar question to this... in order to find the t, you need to find the a and the k values

Answer 4

mmmhhhh ok

Answer 5

T(0) = a + 67 = 190, solve for a

Answer 6

123

Answer 7

T(30) = 123e^10k + 70 = 195, now solve for k

Answer 8

ok is it ok when im done solving it can u check if i got the answer right?? @jtryon

Answer 9

yes, I will check for you

Answer 10

thank you n_n @jtryon

Answer 11

wait, I made an error

Answer 12

ohhh ok

Answer 13

you want to use T(30) = 123e^10k + 67 = 195, because 67 is the surrounding air temp

Answer 14

from the problem

Answer 15

ok

Answer 16

wait im confuse do i subtract 67 from 195 ?? cuz the T(30) is confusing me @jtryon

Answer 17

yes, so you would solve it like any other problem and subtract 67 from both sides

Answer 18

oh okk soo just ignore the T(30)

Answer 19

yes, that is just given information in the problem

Answer 20

The initial temperature of the coffee \(\Large\rm T_o\) is 200.
The ambient temperature \(\Large\rm T_a\) is 67. (I think that's what the a stands for at least..).

And they tell us at time \(\Large\rm t=10\), the temperature has dropped to 195.

\(\Large\rm T(10)=195\).

\[\Large\rm T(t)=T_a+(T_o-T_a)e^{-kt}\]So you can plug some stuff in and solve for k, yah?

Hmm I can't figure out where this T(30) is coming from that you guys are using :(

Answer 21

@zepdrix, I just used that as a placeholder for the equation that I want to solve

Answer 22

but what you are solving is 123e^10k + 67 = 195, for k

Answer 23

Oh :)

Answer 24

ohhh ok then .... getting confusedd

Answer 25

go back to the original equation, you want to find k so that you can plug that in to find t

Answer 26

have you figured out what k is?

Answer 27

Woops*

\[\Large\rm 1\color{red}{3}3e^{\color{red}{-}10k} + 67 = 195\]^I think this is the equation, yes? Hope I did my math correctly >.<

Answer 28

I got 123e

Answer 29

\[123e ^{10k} = 128 \]

Answer 30

thats where i am at ...

Answer 31

It's not going to be 128, less than that

Answer 32

so we know that a = 123

Answer 33

ok

Answer 34

your equation would be 123e^10k + 67 (Ts) = 195 (T0)

Answer 35

hoping you are not confused by T sub-zero and t sub-s

Answer 36

if we subtract 67 from both sides, we get 128 (you are right up to here)

Answer 37

not yet hahaha

Answer 38

I forgot to bring my 1 down

Answer 39

\[\Large\rm T(t)=T_a+(T_o-T_a)e^{-kt}\]\[\Large\rm T(t)=67+(200-67)e^{-kt}\]\[\Large\rm T(t)=67+133e^{-kt}\]
I'm not sure where you got 123 :P
So weird...

Answer 40

what would that equation lead me to ??? @zepdrix

Answer 41

Oh, it is 133 :P

Answer 42

so once you have the a value of 133, you plug that into this equation
133e^10k+67 = 195

Answer 43

and you are solving for the k value

Answer 44

do you know the steps to solve that equation?

Answer 45

\[133e ^{10k} = 128\]

Answer 46

thats as far as i go ...

Answer 47

that is right so far

Answer 48

minus in front of the k, yes? :o

Answer 49

hmm, I don't have a minus in front of the k

Answer 50

the formula is Ae^kt + Ts

Answer 51

@zepdrix, where do you get a minus k?

Answer 52

From the formula he posted :o

Answer 53

what did you get for k, @meln_n

Answer 54

Oh I'm using the formula T= T(t) = ae^kt + Ts

Answer 55

i cant solve it im still stuck at \[133e ^{10k} = 195 \]

Answer 56

Since the coffee temperature is `higher` than the ambient temperature,
I think it makes sense to have a negative k,

as time increases, our temperature falls, getting closer to the ambient temperature.

Answer 57

So you are going to have 133e^-10k = 128

Answer 58

ok

Answer 59

@zepdrix, that look right?

Answer 60

using the formula that he posted

Answer 61

Yah seems like it :D
Solve for k!
You get a really ugly decimal.

Answer 62

Understand how to isolate the k?

Answer 63

You have to divide by 133 on both sides

Answer 64

no i dont thats where i am stuck ... am i suppose to divide next @zepdrix

Answer 65

ohhh ok

Answer 66

Do you know what to do after that?

Answer 67

\[e ^{-10k} = .9624\]

Answer 68

no

Answer 69

Take the natural log of both sides

Answer 70

soooo \[\log e ^{-10k} = \log.9624 \]

Answer 71

would it turn to ln

Answer 72

yes, and then what decimal do you get?

Answer 73

mmmmhhh i got something weird -.038325

Answer 74

what did you plug in?

Answer 75

wait i got .00383

Answer 76

is it still wrong

Answer 77

You're on the right track.
Then you need to divide both sides by -10 to solve for k.

Answer 78

\[\log e ^{-10k} = \log.9624 \]

Answer 79

from that ^^^ i divide -10 ??? how soo

Answer 80

so you have log .9624 / -10

Answer 81

yes i have that !

Answer 82

\[\frac{ \ln.9624 }{ -10 }\]

Answer 83

@zepdrix, I think he has the nasty decimal :P

Answer 84

but when you divide by a negative, you get a positive decimal

Answer 85

Oh sorry, forgot to explain I guess :)

\[\Large\rm \ln(e^{-10k})\]Apply rule of exponents, bring the -10k in front,\[\Large\rm -10k \ln(e)\]Then the ln(e) is simply 1,\[\Large\rm -10k(1)=-10k\]

Answer 86

so what did you get for your k?

Answer 87

i got .00383

Answer 88

So plug your k into your formula:\[\Large\rm T(t)=67+133e^{-\color{orangered}{k}t}\](Or wait til the very end of the problem to plug it in, that might be easier).
They want to know what time value gives us a temperature of 180.\[\Large\rm 180=67+133e^{-\color{orangered}{k}t}\]Solve for t

Answer 89

You have to do some more steps in order to find the time value

Answer 90

Do you know how to get t alone?

Answer 91

not exactly i did something and i got to here

Answer 92

\[e ^{-kt} = .8496\]

Answer 93

So what I have for the equation is 180 = 67 + 133e^-.00383t

Answer 94

ok i used that i just havent input the k value that we found

Answer 95

@zepdrix, so I have 113 = 133e^-.00383t

Answer 96

would it be like this then \[e ^{-.00383t} = .8496\]

Answer 97

what happened to the rest of the equation?