Question

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

Answer 1

This question is about the rate of temperature change, based on surrounding temperature.
Since the thermometer starts in the freezer at 10 degrees, let’s assume that the inside of the freezer is 10 degrees. The room outside the freezer, we are told, is 65 degrees.

Whatever the temperature of the surroundings are, the thermometer will gradually approach this temperature. When the thermometer is moved to the 65 degree room, its temperature increases to approach 65. The greater the temperature difference, the faster this change will be. This also means that the rate of change will not be linear, since the rate of change slows down as the temperatures approach one another.

We are told that it takes 5 minutes, from 1 pm to 1:05, for the thermometer to increase from 10 to 25 degrees. The thermometer stays in the room a bit longer, and increases to some mystery temperature (at least 32) before being placed back in the freezer at some mystery time (before 1:30). By the time 1:30 rolls around, the thermometer has been back in the freezer again, and the temperature has been going back towards the original 10 degrees. The 1:30 snapshot is caught in the middle of this process.
It’s up to us to fill in the blanks of how high the temperature got from outside the freezer and what amount of time that took. This will require us to know the rate of change of the temperature, which we may express in the form of an equation. I don’t mind looking this equation up if needed, but before I do, do you happen to know how we can mathematically represent the rate of temperature change?

Answer 2

The Newton’s law of cooling

Answer 3

SmokeyBrown

Answer 4

@smokeybrown

Answer 5

i feel like if there were answer options that would help picking a better answer -- although i will keep every number as decimal so its as close as possible to the answer.

according to newtons law of cooling whose general equation is \( \frac{dT}{dt}=k(T_o-T_s) \) where \( T_o \) is the temp. of the object (thermometer), and \( T_s \) is the temp. of the surroundings, and \(k\) is going to be the unknown constant.

the solution of it for cooling is going to be \( T=Ce^{kt}+T_s \) where \(C=T_o−T_s\)
the initial temp \( T_o= 10^\circ F \ \ \ ; \ \ \ T_s=65^\circ F \)

\( T=25^\circ F \) at \( t=5 \) min
thus, knowing all that we can find the cnstant \(k\):
\[ 25^\circ F =( 10^\circ F - 65^\circ F ) e^{-k\times 5} + 65^\circ F \]
\[ 25^\circ F = -55^\circ F \times e^{-k \times 5}+65^\circ F \]
\[ -40 ^\circ F = -55^\circ F \times e^{-k \times 5} \]
\[ \frac{-40 ^\circ F }{ -55^\circ F} = e^{-k \times5}\]
\[\ln \left( \frac{-40 ^\circ F }{ -55^\circ F} \right)=\ln\left( e^{-k \times 5}\right) \]
\[ \ln \left( \frac{-40 ^\circ F }{ -55^\circ F} \right)=-k \times 5 \]
\[ \frac{\ln \left( \frac{-40 ^\circ F }{ -55^\circ F} \right)}{5}=-k \]
\[ -0.06369 = -k \]thus:
\[ 0.06369 =k \]
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
now this is where the things get tricky

the moments where the switches happened are unspecified

from \( 1:o5 \) to \( 1:3o \) that would be 25 minutes right?
also, the temperature in a \(65^\circ F \) room drops to \(25^\circ F \) at 1:o5 (assume \(T_1\)) to an unknown temperature (assume \(T_2\)) for an unknown duration (let it be \( t_1\)), that up to \(32^\circ F \)
this would give us a system of 2 equations equations from which we are going to find the value of \( t_2 \) and \( T_2 \)
\[ \begin{cases}
T_{2}=T_{s}+\left(T_{1}-T_{s}\right) e^{-0.06369 \times t_{1}} \ \Rightarrow \ \text{from the room at } \ 65^\circ F \\
T_{3}=T_{s}+\left(T_{2}-T_{s}\right) e^{-0.06369 \times t_{2}} \ \Rightarrow \ \text{from the freezer at } \ 10^\circ F \end{cases} \]
oh and one thing to point out is that \( t_1+t_2=25 \ \ \Rightarrow \ \ t_1=25-t_2 \)
by substituting all the information we have in we would get:
\[ \begin{cases}
T_{2}=65+\left( 25-65\right) e^{-0.06369 \times 25-t_2} \\
32=10+\left(T_{2}-10\right) e^{-0.06369 \times t_{2}}
\end{cases} \]
after you solve it and find the value of both of them, \( T_2 \) is going to be the 2nd part of the answer, so the temperature when the thermometer was taken back, and \( t_2 \) is the time it was taken -- you might have to turn the answer into seconds to be more exact.

you can stop here if you want, and try and solve it yourself, and then check with mine

-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --

now since in the 1st equation the variable on the left side of the equation is an unknown lets work the 2nd one out:

i will treat \( e^{-0.06369 \times t_{2}} \) as a single variable:
\[ 32=10+\left(T_{2}-10\right) e^{-0.06369 \times t_{2}} \]
\[ 32-10=\left(T_{2}-10\right) e^{-0.06369 \times t_{2}} \]
\[ 22= \left(T_{2}-10\right) e^{-0.06369 \times t_{2}} \]
\[ \frac{22}{ \left(T_{2}-10\right)} =e^{-0.06369 \times t_{2}} \]
\[ \frac{ \left(T_{2}-10\right)}{22} =e^{+)0.06369 \times t_{2}} \]
now for the 1st equation:
\[ T_{2}=65+\left( 25-65\right) e^{-0.06369 \times 25-t_2} \]
now we can sparate the exponent of 3, so we can plug in the result from our 2nd equation:
\[ T_{2}=65+\left( 25-65\right) e^{-0.06369 \times 25}\times e^{-0.06369 \times t_2} \]
thus by substituting from earlier you get
\[ T_{2}=65+\left( 25-65\right) e^{-0.06369 \times 25}\times \frac{ \left(T_{2}-10\right)}{22} \]
\[ T_{2}-65=\left( 25-65\right) e^{-0.06369 \times 25}\times \frac{ \left(T_{2}-10\right)}{22} \]
\[ \frac{T_{2}-65}{\left( 25-65\right)}= e^{-0.06369 \times 25}\times \frac{ \left(T_{2}-10\right)}{22} \]
used wolframalpha and got \( T_2=50.14772 \)
back to finding \( t_2 \):
\[ \frac{ \left(T_{2}-10\right)}{22} =e^{(+)0.06369 \times t_{2}}\]
\[ \frac{ \left(50.14772-10\right)}{22} =e^{0.06369 \times t_{2}} \]
\[ 1.82489 =e^{0.06369 \times t_{2}} \]
\[ \ln( 1.82489) = \ln{e^{0.06369 \times t_{2}}} \]
\[ \ln( 1.82489) = 0.06369 \times t_{2} \]
\[ 0.60151 = 0.06369 \times t_{2} \]
\[ \frac{0.60151}{ 0.06369 } = t_2 \]
\[ 9.44433=t_2 \Rightarrow 9" \ 27' \]
now back to our equation where \( t_1=25-t_2 \) ; by plugging what we just got in there we get:
\[ t_1=25"-9" \ 26.7' \]
\[ t_1=15" \ 32.3' \]

so finally we can say that the thermometer was returned back to the freezer 15 minutes and 33 seconds later or at \( (1:05 p.m+00:15":32.3' p.m ) \ \ \) 1:21 minutes and 32 seconds and 3 milliseconds p.m, and its temperature was \(50.14772^\circ F\)