Extra credit problem, due September 10

If you attempt this problem, please write it up separately from the rest of your homework!! Also, please use complete sentences: don't just write down a bunch of formulas and expect me to decipher what you mean by them.

Suppose you pour two very hot cups of coffee, both the same temperature. You add cream to cup 1, then you let them sit for a few minutes, and then you add the same amount of cream to cup 2. Which cup has hotter coffee at this point?

Part (a): What can you figure out just based on common sense? Can you solve the entire problem without using calculus?

Part (b): Now use Newton's Law of Cooling to solve the problem.

Solution:

First, convince yourself that the amount of time the coffee sits before adding cream to cup 2 doesn't make any difference. (This is the part I can't justify, using only common sense.)

If you can get past that first step, then consider the following extreme case: pour the coffee, add cream to cup 1, then let the coffee sit for a week. Brush away any mold that has grown in the interval, then add cream to cup 2. Now it is easy to tell what happens: after a week, both cups of coffee will essentially be at room temperature, so when you add cream to cup 2, the result is:

• if the cream is above room temperature, cup 2 will be hotter;
• if the cream is at room temperature, the temperatures will be the same;
• if the cream is below room temperature, cup 1 will be hotter.

Part (b): We will need to make some assumptions here. [By the way, there may be some complicated chemistry involved here (e.g., the cream may act as an insulating agent, so the coffee with cream may not cool off as fast). I will ignore all such issues and make assumptions that I can actually understand.]

Assumptions: we are starting with the same amounts of coffee, at the same temperature, in each cup, and we add the same amount of cream, at the same temperature, to each. Since we are ignoring chemistry, then we will use Newton's law of cooling with the same cooling constant k for each cup.

We also need to figure out what happens to the temperature of a cup of coffee when we add cream to it. If you have 8 ounces of coffee at 150 degrees, and you add 1 ounce of cream at 40 degrees, it seems reasonable that the temperature of the mixture should be

 (8 x 150  +  1 x 40) / 9  =  8/9 x 150  +  1/9 x 40.

 (1-r)T + rZ.

 (temp of mixture - R) = (1-r)Y + rC.

Notation: since some web browsers don't like superscripts, I will write exp(...) for e raised to the power (...). Let

 A(t) = (temperature of cup 1) - (ambient temperature),
 B(t) = (temperature of cup 2) - (ambient temperature),
 Y    = (initial temperature of coffee) - (ambient temperature),
 C    = (temperature of cream) - (ambient temperature),
 r    = (amount of cream) / (amount of cream + amount of coffee).

Newton's law of cooling: if a body is cooling off, and if its temperature at time t is T(t), then T(t) satisfies the following equation:

 dT/dt = -k (T-R).

 dy/dt = -ky,

 y = y(0) exp(-kt).

So, down to business. Since cup 1 starts at temperature Y above room temperature, then when we add cream to it, we find that its initial value is

 A(0) = (1-r)Y + rC.

 A(t) = [(1-r)Y + rC] exp(-kt).

 B(t) = Y exp(-kt).

 B(t) = (1-r) (current temperature) + rC
      = (1-r) Y exp(-kt) + rC.

 A(t) = [(1-r)Y + rC] exp(-kt) 
      = (1-r) Y exp(-kt) + rC exp(-kt)
 B(t) = (1-r) Y exp(-kt) + rC.

 B(t) - A(t) = rC (1 - exp(-kt)).

 1 > exp(-kt).

• If C is positive (the cream is above room temperature), then B(t) is larger than A(t): cup 2 is hotter.
• If C is zero (the cream is at room temperature), then B(t) = A(t): the cups are the same temperature.
• If C is negative (the cream is below room temperature), then B(t) is smaller: cup 2 is cooler.

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