00:01
Okay, we need to use newton's law of cooling, which is at the derivative of temperature with respect to time is equal to k times whatever temperature it is minus the temperature of the surroundings, which i'm going to call ts.
00:19
That's surrounding temperature.
00:27
There we go.
00:28
Surrounding temperature.
00:30
So we need to solve this differential equation to be able to extract a function of temperature with respect to times.
00:39
So to do that, i'm going to divide both sides by t minus t s to get one over t minus ts.
00:46
Well, first of all, we know that ts is 35, so why don't i replace this with a 35? and then i get, excuse me, there's a d, dt, dt here, is equal to k.
01:04
There we go.
01:06
Okay, so this differential equation is what we call separable.
01:10
It means that if i integrate both sides with respect to t with respect to, a little t, i should say, then i end up getting 1 over t minus 35 d big t is equal to k d t, little t, just like that.
01:30
So one over big t minus 35 d big t is equal to kd little t.
01:36
And this is good because now i can integrate both sides with respect to their own variable.
01:44
And that's going to give me the ln of t minus 35.
01:49
Is equal to k t little t k little t now or i forgot something i forgot the arbitrary constant there we go now i'm going to exponentiate both sides to get that t minus 35 is equal to e to the k little t plus c like that which is equal to e to the c times e to the k t now e to the c since c is just an arbitrary constant, this is just some other arbitrary constant a.
02:31
So i get that t minus 35 is equal to a, e to the kt.
02:40
So a function of temperature with respect to time would be a, e to the k little t plus 35.
02:51
And now i can use the other information to figure out what a is and to figure out what k is.
02:56
Once i do that, i have an expression that will tell me the temperature at any time.
02:59
And t of the water.
03:01
So to figure out what a is, i plug in the initial temperature.
03:07
So the initial temperature at 2 p .m.
03:09
Was 18 degrees.
03:11
That's going to equal a, e to the k times zero plus 35, which is going to give me 18 is equal to a plus 35.
03:22
So a must be negative 17 when i subtract 35 on those sides.
03:29
So t, little t, is equal to nciful.
03:33
Negative 17 e to the kt, k little t plus 35.
03:42
Now, to solve for k, i can use the other piece of information.
03:47
It says that after 25 minutes, the water's temperature was 26 degrees.
04:00
So this is what we have now.
04:01
I'm going to subtract 35 from both sides to get negative 9.
04:04
It's equal to negative 17, e to the kt.
04:08
I'm going to divide both sides by negative 17 to get 9 over 17...