 # An experiment to determine coffees cooling rate on three different conditions

## Other differential equations

All the solutions grow exponentially All the solutions decay to zero With this in mind, let us examine another realization of a differential equation, this time connected with the property of cooling or loss of heat of a warm object in a colder environment. The following "Law" is an approximate description of experimentally observed behaviour. Newton's Law of Cooling Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature i. Newton's Law makes a statement about an instantaneous rate of change of the temperature. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation.

1. Taking logarithms of both sides, we find that Thus, using the fact that we have Thus, it will take a little over half an hour for Jim's soup to cool off enough to be put into the refrigerator.
2. Here a bit of care is needed.
3. Would Newton's Law of Cooling apply just as before?

The solution to this equation will then be a function that tracks the complete record of the temperature over time. Newton's Law would enable us to solve the following problem.

## The Coffee Cooling Problem

The Big Pot of Soup As part of his summer job at a resturant, Jim learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. He also found out that, while refrigeration was essential to preserve the soup overnight, the soup was too hot to be put directly into the fridge when it was ready.

The soup had just boiled at 100 degrees C, and the fridge was not powerful enough to accomodate a big pot of soup if it was any warmer than 20 degrees C. Jim discovered that by cooling the pot in a sink full of cold water, kept running, so that its temperature was roughly constant at 5 degrees C and stirring occasionally, he could bring tht temperature of the soup to 60 degrees C in ten minutes.

How long before closing time should the soup be ready so that Jim could put it in the fridge and leave on time?

1. We have just seen yet another example of a simple differential equation and how it can be used to make predictions.
2. We need to wait until , so at that time.
3. Will this term be increasing or decreasing with time?
4. In particular, what will happen to the term?

Let us summarize the information briefly and define notation for this problem. The rate of change of the temperatureis by Newton's Law of Cooling proportional to the difference between the temperature of the soup and the ambient temperature This means that: Here a bit of care is needed: Clearly if the soup is hotter than the water in the sinkthen the soup is cooling down which means that the derivative should be negative.

Remember the connection between a decreasing function and the sign of the derivative? This means that the equation we need has to have the following sign pattern: The independent variable is for time, the function we want to find isand the quantities are constants.

In fact, from Jim's measurements, we know thatbut we still don't know what value to put in for the constant. We will discuss this further below. Can it also describe the case of a cold object heating up in a warmer environment? Back to the same old equation The equation we arrived at above looks different from the ones we have just investigated, but as we shall soon see, the difference is rather superficial.

Indeed, by defining a new variable, we will show that the equation is really completely related to the exponential decay seen previously. What a nice surprize! By defining this new variable, we have arrived once more at the familiar equation whose solution is well known to us, namely: We can use this result to conclude by plugging in and that It follows that We found the solution in general form, but it looks quite complicated.

Let's try to understand this expression and its predictions in the case of the problem described above. How the soup will cool From the information in the problem, we know that so that, We also know that after 10 minutes, the soup cools to 60 degrees, so that.

Plugging into the last equation, we find that Rearranging, The steps are much the same as in our previous work in the example on radioactive decay. In the last step we took a reciprocal of both sides of the equation. This just makes all the quantities come out to be positive in the next step, so it is done for convenience, though it is not an essential step. We have found that Taking the natural an experiment to determine coffees cooling rate on three different conditions of both sides, and solving forwe find that Thus, So we see that the constant which governs the rate of cooling is per minute.

Now we can specify the solution fully, since all constants have been determined from the information in the problem. The prediction is that the temperature of the pot of soup at time t will be The behaviour of this solution is shown in the diagram. What will happen after a very long time?

What will the eventual temperature of the soup be? In particular, what will happen to the term? Will this term be increasing or decreasing with time? Would it still cool the same way? Would Newton's Law of Cooling apply just as before? Why are we assuming that the pot is well-stirred? How would this change our assumption that the ambient temperature was a constant? Solving Jim's Soup problem To finish our work, let us determine how long it takes for the soup to be cool enough to put into the refrigerator. We need to wait untilso at that time: This equation can be solved for in much the same way as before. Subtracting 5 from both sides and dividing by 95 we get: Taking logarithms of both sides, we find that Thus, using the fact that we have Thus, it will take a little over half an hour for Jim's soup to cool off enough to be put into the refrigerator.

We have just seen yet another example of a simple differential equation and how it can be used to make predictions. To summarize what we found, here is the connection between the differential equation of Newton's Law of Cooling and its solution: Newton's Law of Cooling. 