00:01
Hello and welcome to this problem.
00:02
You were asked two questions based on an object that is cooling and a given room.
00:10
The first question is to find the temperature of the object 20 minutes in, and the second is to find the time required for the object to cool down to 55 degrees celsius.
00:21
Some additional information that we're given is that the object initially starts at 25 degrees celsius, the temperature at 10 minutes is 65 degrees celsius, the environment is 25 degrees celsius, and the object cools at a rate of k theta.
00:40
So we'll have d -d -theta d -teta -d -t is equal to negative k -theta, that is per the problem, where theta is the difference between the current temperature and the temperature of the environment.
01:02
So theta is equal to t minus t environment.
01:08
Great, so let's solve this.
01:10
We can separate and solve this very ordinary differential equation.
01:15
And since this is such a common differential equation, i'll just write the solution.
01:19
It's of the form theta equals c, e to the minus kt.
01:26
All right, so c is a constant here.
01:30
And let's solve for this constant.
01:33
So we know that initially, at t of zero, the temperature is 75.
01:38
So we'll have theta is equal to 75 minus 25.
01:41
That's the environment.
01:43
So we'll have 50 is equal to c.
01:47
E to the 0 is just c.
01:49
So c is 50.
01:55
Great.
01:56
And now we can solve for k.
02:00
That's an important value for this.
02:02
If we're going to figure out the temperature of 20 minutes.
02:06
So now that we know c, we can plug in the value at t equals 10.
02:12
So we'll have theta is equal to 65 degrees minus 25 degrees, which is 30.
02:22
And this will, right, yes.
02:29
Nope, no, this is 40.
02:31
So looked wrong.
02:33
So it's 40.
02:35
And therefore, we can plug the same.
02:37
In as 40 is equal to 50, e to minus k times 10.
02:45
This is 10 minutes in times 10.
02:49
We can solve this then.
02:50
This will be 4 over 5 natural log is equal to negative k times 10.
02:59
And we'll finish this off as k is equal to the natural log of 4 over 5.
03:13
Yeah, over 10.
03:17
Looks good.
03:18
All right.
03:20
This k drops because we are just solving for k...