00:01
Probably have a pizza taken out of a 450 degree oven and set into a 70 degree room.
00:06
After five minutes, its temperature is 300, and we want to know when will it be 135, and how long until it's 160? and after a long time, will it's temperature be? well, this is a cooling problem, so we're going to use newton's law of cooling.
00:20
It's right here.
00:22
Little t, that's the time.
00:25
U of t, that's the temperature at time t.
00:30
Big t is the temperature of the room.
00:32
Room or the water, wherever the thing is that's cooling.
00:35
It's a plus sign right here.
00:38
U -subs -0, that's the initial temperature of the object.
00:42
So here it's the pizza and it's going to be 450 because that's how much it is when it's coming out in the oven.
00:48
And then k is the cooling constant and it will be a negative number.
00:54
Okay, so 450, that's the initial temperature of the pizza.
00:58
So that's going to go right there.
01:01
And then 70 degrees, that's the temperature of the room.
01:03
That's going to go there and there.
01:07
And then after five minutes, a temperature will be 300.
01:11
Okay, well, let's do the first two, and then we'll come back to that.
01:15
So u of t is 70 plus 450 minus 70, e to the kt.
01:25
So that's 70 plus 380, e to the kt.
01:31
Okay, now that next sentence is to find k for us.
01:35
It's giving us a time and a temperature.
01:38
So we'll plug those in.
01:40
The temperature is 300 when the time is 5.
01:46
So e to the 5k.
01:49
So we're trying to solve for k first.
01:52
So subtract 70, divide by 380.
02:06
Now the only way to get that exponent down is to take the logarithm of both sides because that is the inverse function of raising a to the power.
02:19
So when you do ln of e to the 5k, the ln in e cancel, you get 5k.
02:25
Last step, divide by five...