Question

if temperature of a system decreases from 60c to 50c in 10 mins and then 50c to 42c in next 10min.what is the temperature of surrounding.

Answer 1

(My apologies for the delayed answers, but first my browser crashed when I was in middle of typing and then when I opened up the group chat, the editor was cleared again! Also, sorry that I couldn't figure out how to stop the editor from putting all math expressions on a new line. I'll fix this later if possible.)

According to Newton's cooling law, the rate of change of temperature is directly proportionate to the difference in temperatures, that is to say: \[\frac {dT}{dt}=k(T-T_0)\], where \[T=T(t)\] is the time-dependant temperature of the cooling object and \[T_0\] is the temperature of the environment, assumed to be constant. k is the proportionality constant which can be calculated from experimental data.

We solve the separable differential equation explicitly:
\[\frac {dT}{dt} = k(T-T_0) \]
\[\frac {dT}{T-T_0}=k \]
\[\int \frac {dT}{T-T_0}dt=\int k dt \]
\[\ln(T-T_0)=kt+C_1\]
\[T-T_0=e^{kt+C_1}\]
\[T(t)=e^{kt}e^{C_1}+T_0\]
\[T(t)=Ce^{kt}+T_0\]

To have a complete, although idealised, model for the given system, we need to define the constants C, k and finally the requested \[T_0\].

The temperature of the object in the beginning of the experiment is 60 degrees Celsius, so we can write down the equation \[T(0)=60\] or \[Ce^{k \cdot 0}+T_0=60\]. This gives us a relation between C and \[T_0\]:
\[C+T_0=60\].

After ten minutes, \[T(10)=50\] or \[Ce^{k \cdot 10}+T_0=50\].
After ten more minutes at \[t=20\], we have \[Ce^{k \cdot 20}+T_0=42\].

We have obtained the following well-defined system of three equations:
\[\begin{cases} C+T_0=60 \\ Ce^{10k}+T_0=50 \\Ce^{20k}+T_0=42 \end{cases}\]

We can proceed (for example) by solving for \[C\] in the first equation. We have \[C=60-T_0\]. Substituting this to the second equation, we get \[(60-T_0)e^{10k}+T_0=50\].

We then solve k from the third equation:
\[Ce^{20k}+T_0=42\]
\[e^{20k}=\dfrac{42-T_0}{C}\]
Taking square root from both sides of the equation we obtain:
\[e^{10k}=\sqrt{\dfrac{42-T_0}{C}}\].
Substituing in \[C=60-T_0\] we get
\[e^{10k}=\sqrt{\dfrac{42-T_0}{60-T_0}}\]

This whole expression can now be substituted for \[e^{10k}\] in our previous equation for \[T_0\] and k.

\[(60-T_0)\sqrt{\dfrac{42-T_0}{60-T_0}}+T_0=50\]
Moving \[T_0\] to the other side and squaring, we get:
\[((60-T_0)\sqrt{\dfrac{42-T_0}{60-T_0}})^{2}=(50-T_0)^{2}\]
\[(60-T_0)^{2} \dfrac{42-T_0}{60-T_0}=(50-T_0)^{2}\]
Which simplifies further to:
\[(60-T_0)(42-T_0)=(50-T_0)^{2}\]
or
\[(T_0-50)^{2}-(T_0-60)(T_0-42)=0\]
which by polynomial expansion gives:
\[T_0^{2}-100T_0+2500-T_0^{2}-42T_0-60T_0+2520=0\]
or
\[-202T_0=-5020\]
And, finally:
\[T_0=\dfrac{-5020}{-202}=\dfrac{2510}{101}\approx25\] (degrees Celsius).

Answer 2

@ joonasD6 ..thank u so much ...this is really cool..

Answer 3

Cheers. :) Do tell if I should have more steps visible.

Answer 4

its very clear...thanks a lot.. i was breaking my head over this from yesterday..

Answer 5

Depending on the order and method of solving the system of equations, it can turn out very ugly. :)

Answer 6

ya thats really true..actually i am life science student ..its been long since i studied physics in high school..but the explanation u gave above is just awesome..i really feel like studying physics again....

Answer 7

Kind of sad that part of the joy of physics can sometimes drown in problems in maths. But Newton's law of cooling is a good example of how experimental results turn to useful mathematical models.

If you solved C and k there as well, you'd then have the "full" model with what you can figure out (just by plugging in desired values for time) what the temperature is at any given moment or what it was even before the experiment started! (Negative values of t.) Or the other way around: how long does it take for the object to reach a certain temperature. I actually did model something similar some time ago, to estimate how long did it take to make ice cubes in my freezer. :)

Answer 8

@JoonasD6 your physics is great and the answer is awesome, but when I solve the set of equations with WolframAlpha, I get T_0=10, not 25:
http://www.wolframalpha.com/input/?i=solve+ {c%2Bt_0%3D60%2Cc*e^%2810*k%29%2Bt_0%3D50%2Cc*e^%2820*k%29%2Bt_0%3D42}

I can't immediately see where the mistake is. :-(

(Openstudy seems to break the link, but copy-paste it to the browser.)

Answer 9

Indeed, apparently the solution T_0=25 doesn't solve the equations...
http://www.wolframalpha.com/input/?i=solve+ {c%2B25%3D60%2Cc*e^%2810*k%29%2B25%3D50%2Cc*e^%2820*k%29%2B25%3D42}

Wonder where I did a mistake. (The method of solution is still valid.)

Answer 10

Ah, found it. I missed the minus sign when expanding the squared binomials. :)

Answer 11

wel a lot of thanks to u both...but i just got the answer for this question and it was 20 degree c...they didnt solve this but just gave the answer...so was wondering whether the answer they gave is wrong??

Answer 12

Always ask for the full answer. Otherwise there is no way to check if it is right or wrong.

One option might be that the teacher uses some other model for cooling down.

Answer 13

There indeed is a big difference (education-wise) between the answer and the solution.

Answer 14

hey its 10degre only ...bloody... they just gave the solution ...there was mathematical error in the last step.. so the answer was wrong...thanks a lot to u both.. u both really rock...

Answer 15

I do have to add that this was a very well rigged exercise. It's not often you get an exponential model and end up working with fractions and integers!