A freshly brewed cup of coffee has temperature $ 95^o C $ in a $ 20^o C $ room. When its temperature is $ 70^o C, $ it is cooling at a rate of $ l^o C $ per minute. When does this occur?
here we have a Newton's law of cooling problem, and the Newton's law of cooling is based on the same kind of exponential function relationship as all the others in this section that we've learned and the background of that is a differential equation. The rate of change is proportional to the Y value. Do I d. T equals K times? Why? So we're told the rate of change when the temperature of the coffee is 70 degrees is one degree per minute. And since the temperature of the coffee is going down, we're going to call that a negative one. So what we can do is we can say that negative one is equal to K times Why, and we can sell for K so negative one equals k times. Why now, in our Newton's law of cooling formula, we have replaced the why with t minus T's of s, where T is the temperature at a given time and T's of s is the temperature of the room, which in this case is 20 degrees. So we have negative one equals K times 70 the temperature of the time minus 20 the temperature of the room. So have negative one equals K times 50. So we're going to divide both sides of that by 50 and we get K is negative 0.2 So we can put that into our model for Newton's Law of cooling for this problem. So we have the temperature of the object t minus the temperature of the surroundings. 20 is equal to the initial temperature of the object, which we're told is 95 degrees, three initial temperature of the coffee minus the temperature of the surroundings 20 multiplied by each raised to the k times t. So we just found Kay. So here's our model that we're going to use, and what we want to know is, how long does it take for the coffee temperature to get to 70? All right, so here's our model again. We're putting 70 and for the temperature of the coffee and we're solving for the time, 70 minus 20 is 50. We're going to divide both sides of the equation by 75. That gives us 2/3. They were going to take the natural log of both sides, so we have the natural log of 2/3 equals negative 0.2 temps tea, and then we'll divide both sides of the equation by negative 0.2 And let's approximate this with our calculators, I would get approximately 20.27 minutes. That's the amount of time it would take for the coffee to go from its original 95 degree temperature down to 70 degrees.