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

So you answered your own question. makes the job much easier for us now doesn't it? :D

Answer 2

That looks like a physics question. We might need some associated thermal formulas. I have not had a heat ransfer course in years so I am a little rusty in this field. What course is this question for?

Answer 3

looks like some sort of newton's law of cooling puzzle

Answer 4

There is an answer given to us but i don't know the solution. im stuck at the point where the temp is heated and cooled after.

Answer 5

It is a newtons law of cooling problem

Answer 6

@znerte Are you studying law of exponential change?

Answer 7

yes

Answer 8

so far ive got the k for this question which is -.0637

Answer 9

lets see
the temperature difference starts at 55 degrees at 1:00
five minutes later it is 40 degrees, so you can model this with \[A(t)=55\left(\frac{8}{11}\right)^{\frac{t}{5}}\]

Answer 10

or you can also do it by using the (rather slavish) formula
\[55e^{whatever}\]

Answer 11

now i am kind of stuck, but i guess we are to assume that the freezer is a constant temperature of ten degrees, right?

Answer 12

Good assumption @satellite73

Answer 13

i think so because it is at 10F when it is removed from the freezer

Answer 14

lol i am a bit slow this morning
so far all i have is the temperature of the thermometer outside of the room is \[A(t)=65-55e^{-.067t}\]

Answer 15

i meant outside of the freezer

Answer 16

Basically we apply Newton's Law of Cooling as mentioned earlier. We can derive the formula using separable DEs/law of exponential change but I will choose not to do that.

Going directly to the problem, we know that the initial temperature inside the freezer at 1 pm was 10F. It was then placed inside the room which was at temperature 65F and at 1:05 pm, 5min later, the thermometer reads 25F. From this data here and using Newton's Law of Cooling, we get:\[\bf T-T_s=(T_0-T_s)e^{-kt} \rightarrow 25-65=(10-65)e^{-k(5)}\]\[\bf \implies -40=(-55)e^{-5k} \implies e^{-5k}=\frac{ 8 }{ 11 }\]\[\bf \implies -5k = \ln \left( \frac{ 8 }{ 11 } \right) \implies k=\ln \left( \frac{ 11 }{ 8 } \right)^{\frac{1}{5}} \approx -0.0637\]
Good so far? @znerte

Answer 17

he/she wrote that above

Answer 18

I shall move on then.

Answer 19

yeah please, because i can figure out the rate of warming, but not the rate of cooling
want to see how you do it

Answer 20

question is the k is the same at the rate of cooling

Answer 21

why would it be?

Answer 22

ok lets assume that it is the same
what equation can we solve?

Answer 23

it is out for \(t\) minutes and back in for \(30-t\) minutes

Answer 24

holy crap i think i got it!

Answer 25

lets see if this makes sense
we assume the rate of cooling is the same as the rate of heating
since i know literally nothing about physics lets just assume it for a second okay?

Answer 26

you have found the rate at \(-.0637\) so the equation for the temperature of the thermometer outside the freezer is
\[65-55e^{-.0637t}\]

Answer 27

at some time \(t_0\) it is put in the freezer and the temperature is
\[65-55e^{-.0637t_0}\] i just used \(t_0\) to indicate that the above is some constant that we do not know
now assuming the rate hasn't changed, the temperature once it is back in the freezer is
\[\left(65-55e^{-.0637t_0}\right)e^{-.0637t}+10\]

Answer 28

out for \(t_0\) minutes, then back in for \(30-t_0\) minutes

Answer 29

\[\ln \frac{ 32-10 }{ 65-55^{-.0637t}-10 }=-0.0637(30)+0.0637t\]

Answer 30

lets try solving
\[\left(65-55e^{-.0637t_0}\right)e^{-.0637(30-t_0)}+10=32\]

Answer 31

solving for t i get 20.555

Answer 32

then go with that, because my answer is not right !