Question

Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. So the rate of cooling for a bottle of lemonade at a room temperature of 75°F which is placed into a refrigerator with temperature of 38°F can be modeled by dT/dt=k(T-38) where T(t) is the temperature of the lemonade after t minutes and T(0) = 75. After 30 minutes the lemonade has cooled to 60°F, so T(30) = 60.

To the nearest degree, what is the temperature of the lemonade after an additional 30 minutes?

Answer 1

dT/dt=k(T-38)

This is a separable differential equation.

\[\frac{dT}{dt}=k(T-38) ----->\frac{1}{(T-38)}dT=k*dt\]
Integrate both sides:
\[\int\limits_{}^{} \frac{1}{(T-38)}dT=\int\limits_{}^{} k*dt\]
\[\ln(T-38)=k*t +constant\]

Answer 2

undo the ln with e

Answer 3

\[T-38=e^{k*t +constant}\]

Answer 4

\[T-38=Ae^{k*t}\]
A is just the constant pulled out of the exponent

Answer 5

\[T=38+Ae^{k*t}\]

Answer 6

T(0) = 75
T(30) = 60

Answer 7

I don't feel like calculating what A and k are, but it should be doable.

Answer 8

@Michele_Laino can u help me from here

Answer 9

@Michele_Laino

Answer 10

here we have to solve the differential equation, first.
In order to that, I will apply the formula, which I wrote in my tutorial

Answer 11

ok

Answer 12

here is my tutorial, for more explanation:
http://openstudy.com/study#/updates/552b4f50e4b04e5707c1d380

Answer 13

@Michele_Laino okay. i will be back in about 35 min. i have to go to lunch rn. but i promise ill be back.

Answer 14

ok!

Answer 15

@Michele_Laino im back

Answer 16

I can rewrite your equation as below:
\[\Large \frac{{dT}}{{dt}} - KT = - 38K\]
so, comparing that formula, with the general formula of my tutorial, I get:
\[\Large \begin{gathered}
P\left( t \right) = - K,\quad Q\left( t \right) = - 38K \hfill \\
\hfill \\
{e^{\int P }} = {e^{ - Kt}},\quad {e^{ - \int P }} = {e^{Kt}} \hfill \\
\end{gathered} \]
then the solution, is:
\[\Large T\left( t \right) = {e^{Kt}}\left( {C + \int { - 38K\;{e^{ - Kt}}dt\;} } \right)\]
and finally:
\[\Large T\left( t \right) = C{e^{Kt}} + 38\]

Answer 17

okay so how do I determine the temp after 30 min? @Michele_Laino

Answer 18

we have to set T(0)=75, so we get:
\[75 = T\left( 0 \right) = C{e^{K \times 0}} + 38 = C + 38\]

Answer 19

so c=37?

Answer 20

yes!
Nevertheless, our function T(t) is an increasing function

Answer 21

I think there is a typo error in the text of the exercise

Answer 22

heres the screenshot

Answer 23

I think that the rate of cooling, has to be this:
\[\frac{{dT}}{{dt}} = - K\left( {T - 38} \right)\]
where K>0

Answer 24

otherwise, the formula of T above is correct, if K<0

Answer 25

Let's see if that conjecture is true

Answer 26

we have:
\[T\left( t \right) = 37{e^{Kt}} + 38\]
now I substitute the condition:
T(30)=60, and I can write:
\[60 = T\left( {30} \right) = 37{e^{30K}} + 38\]
and:
\[{e^{30K}} = \frac{{60 - 38}}{{37}} = ...?\]

Answer 27

@familyguymath

Answer 28

0.59

Answer 29

ok! better is 22/37

Answer 30

so we have:
\[\Large {e^{30K}} = \frac{{22}}{{37}}\]

Answer 31

right

Answer 32

now, we have to compute this:
T(30+30)=...?

Answer 33

namely:
T(60)=...?

Answer 34

we can write:
\[\Large \begin{gathered}
T\left( {60} \right) = 37{e^{60K}} + 38 = 37{\left( {{e^{30K}}} \right)^2} + 38 = \hfill \\
\hfill \\
= 37 \times {\left( {\frac{{22}}{{37}}} \right)^2} + 38 = ...? \hfill \\
\end{gathered} \]

Answer 35

@familyguymath

Answer 36

give me a sec

Answer 37

51.08

Answer 38

or 1890/37

Answer 39

ok! That's right!

Answer 40

T(60)=51.08 °F

Answer 41

alright so what about the second half?

Answer 42

we have to solve this equation:
\[\Large 55 = 37{e^{K{t_0}}} + 38\]
and we have to compute t_0

Answer 43

so, we can write:
\[\Large {e^{K{t_0}}} = \frac{{55 - 38}}{{37}} = ...?\]

Answer 44

17/37

Answer 45

ok!

Answer 46

now, I can write this:
\[\Large {e^{\left( {30K} \right) \times \left( {\frac{{{t_0}}}{{30}}} \right)}} = {\left( {{e^{30K}}} \right)^{{t_0}/30}} = {\left( {\frac{{22}}{{37}}} \right)^{{t_0}/30}} = \frac{{17}}{{37}}\]

Answer 47

so:
\[\Large {t_0} = \frac{{\log \left( {17/37} \right)}}{{\log \left( {22/37} \right)}} = \frac{{\log 17 - \log 37}}{{\log 22 - \log 37}} = ...?\]

Answer 48

@familyguymath

Answer 49

1.49594

Answer 50

yes! Nevertheless I compute another time t_0

Answer 51

I try to check my computation

Answer 52

okay and?

Answer 53

sorry, the right formula, is:
\[\Large {t_0} = 30 \times \frac{{\log \left( {17/37} \right)}}{{\log \left( {22/37} \right)}} = 30 \times \frac{{\log 17 - \log 37}}{{\log 22 - \log 37}} = ...?\]

Answer 54

@familyguymath

Answer 55

45 @Michele_Laino

Answer 56

that's right!