Question

at 1:00 pm a thermometer reading 10F is removed from a freezer and placed in a room whose temperature is 65 F. at 1:05 pm, the thermometer reads 25 F. Later, the thermometer is placed back in the freezer. at 1:30 pm the thermometer reads 32 F. when was the thermometer returned to the freezer and what was the thermometer reading at that time? Answer: 1:20 pm; 50.15 F

Answer 1

the formula is
\[\huge \ln (\frac{T-T_m}{T_o - T_m}) = kt\]

Answer 2

where T = final temp
T_o = initial temp
T_m = outside temp
k = constant
t= time

Answer 3

can you do it @nbouscal ?

Answer 4

How can the answer be that it was reading 50.15 F at 1:30 pm if the question says that it read 32 F at 1:30 pm?

Answer 5

oops typo..that's 1:20

Answer 6

the answer should be 1:20

Answer 7

Hrm. No, I do not know how to solve this problem.

Answer 8

aw darn :(

Answer 9

@Limitless can you?

Answer 10

\[k=\frac{\ln\left(\frac{T-T_m}{T_0-T_m}\right)}{t}\]
\(T_0=10\), \(T=25\), \(T_m=65\), \(t=1:05-1:00=5\)
\[k=\frac{\ln(40/55)}{5}=\frac{\ln(8/11)}{5}\]

I'm stuck at reasoning out how to pinpoint the position in the gap of 1:05 to 1:30.

Answer 11

i think i got an equation differen from this one?

Answer 12

lol nevermind i have no idea what to do here

Answer 13

so k = -0.0637?

Answer 14

If my logic is correct and you used the appropriate calculator, yes.

Answer 15

I don't get exactly what you have written as the answer

Answer 16

i get 1:20.555 and a temp of 50.147

Answer 17

@Zarkon, that's part of the answer... I think.

Answer 18

I did not write a complete answer; I just posted my thoughts so far.

Answer 19

Talking about answer in the OP

Answer 20

Oh, okay.

Answer 21

It's rounding, by the way.

Answer 22

I made a couple assumptions to do it though

Answer 23

I assumed the freezer temp was 10F. and so k was the same for growth and decay

Answer 24

That is a valid assumption, I think.

Answer 25

And I used my calculator to solve the equations :)

too lazy to do by havnd

Answer 26

*hand

Answer 27

yes..i used calculator too..but i think i got 8 or something

Answer 28

could you post the ln you got? i think that's where i went wrong

Answer 29

so I get \[T(t)=65-55\left(\frac{8}{11}\right)^{t/5}\]

my calculator simplified it to this....for the first part

Answer 30

then I made the equation

\[T_2(t,s,w)=10+(s-10)e^{-(t-w)\frac{\ln(11/8)}{5}}\]

then I solved \[T_2(30,T(t),t)=32\] for \(t\) ;)

Answer 31

got \[t=20.5554814377\]

Answer 32

uhmm maybe i should post my solution and you can spot where i went wrong...

Answer 33

because i cannot see my mistake from your solution since beginning is different

Answer 34

and \[T(\text{the above})=50.1479302375\]

Answer 35

my \(T(t)\) can be written as

\[T(t)=65-55e^{-t\frac{\ln(11/8)}{5}}\]

Answer 36

which I believe should be the same as Limitless

Answer 37

I'm out of pepsi now so my powers might be fading

Answer 38

ugh i dont understand this T_T

Answer 39

I don't know if I can eloquently describe what I did

Answer 40

we have two main issues here...we dont know the time it went back into the freezer and thus we dont know the temperature

Answer 41

okay i set up two conditions

first condition T_m is 65

second contition T_m = 10

is that right

Answer 42

my \[T_2(t,s,w)\]

has 3 variables...the first t is the time at the end which will be 30
the S is the temperature at time (t) (unknown...it is \(T(t)\)

and w is the time at which we put it back in the frezer

Answer 43

hmm i see

Answer 44

t is the time before 30?

Answer 45

we then know that \[T_2(30,T(t),t)=32\]

since this is the temp at time 30

Answer 46

yes

Answer 47

wait...ill try to re-solve

Answer 48

i quit..i cant do this...sorry...and thanks for all your help

Answer 49

sorry I couldn't help you more