Question

the initial temp of an beer is 70 degrees, but it in a 10 degree freezer
1/2 a minute later it is only 50 degrees

Answer 1

work with the difference in temperatures, that is what decays

Answer 2

70-10=?

Answer 3

60

Answer 4

so initial difference is 60, like having (0,60) on the graph

Answer 5

right i understand this part

Answer 6

then in half a minute, what is the difference in temps?

Answer 7

10?

Answer 8

jeez read the damned problem

Answer 9

in the book i mean, just read what it says in the book

Answer 10

50

Answer 11

NO KEEP READING

Answer 12

i am

Answer 13

15?

Answer 14

A thermometer is removed from a room where the
temperature is 70° F and is taken outside, where the air
temperature is 10° F. After one-half minute the ther-
mometer reads 50° F.

Answer 15

the initial difference it 70 - 10 = 60
after half a minute the difference is 50 - 10 = ?

Answer 16

OH

Answer 17

oh...

Answer 18

I don't like the way they worded it

Answer 19

too bad
we are not done

Answer 20

ok

Answer 21

we have (0,60) and (.5,40) we need the exponential model for that

Answer 22

\[T=60e^{kt}\] where if \[t=.5, T=40\] solve \[40=60e^{.5k}\] for \[k\]

Answer 23

ln(.2)/.5

Answer 24

wait

Answer 25

.66

Answer 26

\[40\div 60\neq .2\]

Answer 27

ln(.66)/.5

Answer 28

ok good enough
now be careful

Answer 29

those are the temperature DIFFERENCES
the actual temp you have to add the 10 degree ambient temperature
so it really should be \[E=60e^{-.118t}+10\]

Answer 30

so because there is a difference between the actual and whatever they give you, you have to add the difference?

Answer 31

you have to add the ambient temp, it is the differences that decays

Answer 32

oh ok

Answer 33

the temp of the heated object decays to the ambient temp

Answer 34

now if you want the diffeq solution, which is identical, start with \[\frac{dT}{dt}=k(T-10)\]

Answer 35

ok

Answer 36

separate get \[\frac{dT}{T-10}=kdt\] so \[\int \frac{dT}{T-10}=\int kdt\] making \[\ln(T-10)=kt\] and therefore \[T-10=Ce^{kt}\]

Answer 37

but it is identical only with the +10 out at the end
the initial difference it 60 so C = 60 and you still find k the same way

Answer 38

T = Ce^(kt) + 10

Answer 39

This is way easier than the other way

Answer 40

identical

Answer 41

easier

Answer 42

identical

Answer 43

and now i am done
try one yourself, see that they are not that hard
i think the mixture ones might use an integrating factor, not sure they separate

Answer 44

I will tomorrow. I have to get to bed

Answer 45

Thank you so much!

Answer 46

read example in 1.3 of setting up the mixture
then read example of that same problem in 3.1

Answer 47

good night!

Answer 48

yw

Answer 49

okie dokie sounds good! Goodnight!