The Coffee Cooling Problem  

Mary Ellen Verona
Maryland Virtual High School of Science and Mathematics
Silver Spring, Maryland 20901
Based on a similar problem from the 2nd Scientific American Book of Mathematical Puzzles and Diversions by Martin Gardner

It's your planning period and you've just poured yourself a cup of coffee and taken cream from the refrigerator when the fire alarm rings. Should you add the cream to the coffee now or wait till you get back? Which method will result in a warmer cup of coffee upon your return? The answer to this problem is not intuitive to most of us. MVHS students were asked to solve this problem under the following conditions:

Given 250 grams of coffee at 90 degrees Celsius and 50 grams of cream at 10 degrees Celsius in a room whose temperature is 20 degrees Celsius.

Case 1: Cream is mixed with coffee immediately, creamed coffee held for 30 minutes.
Case 2: Cream and coffee are held at room temperature for 30 minutes and then mixed.

In which case is the creamed coffee hotter? What is the temperature in each case? Assume that the temperature of the black coffee will be approximately 52 degrees after 30 minutes at room temperature.

Note: Cooling is inversely proportional to mass and directly proportional to the temperature difference between the substance and its environment. It is assumed that 1 gram of coffee, cream, and creamed coffee at the same temperature will cool at the same rate (have the same specific heat). It is not necessary that you understand specific heat to solve this problem. The given temperature of black coffee after 30 minutes determines a missing factor.

Students from six high schools submitted their answers. Two high schools took an experimental approach and used CBL's to find that the mixed coffee was hotter. But the difference in coffee temperature at one school was 7 degrees Celsius less than the difference at the other school. What outside factors might account for this difference?

Students at the other schools tried modeling the problem. First, they used a weighted average to determine the initial temperature of the creamed coffee. Then they found the cooling rate for the black coffee by using the given information that black coffee would drop from 90 to 52 degrees Celsius in 30 minutes. They applied this same rate to the cream and creamed coffee in spite of the differences in mass. This error produced an incorrect answer.

Modeling software, such as STELLA or a spreadsheet, can be used to model this problem effectively. In the STELLA model shown, the cream, creamed coffee, and coffee stocks are initialized to 10, 87.67, and 90 degrees Celsius, respectively. Air temp is held constant at 20 degrees Celsius because the volume of air in the room is so much greater than the volume of the liquids. At each time step, the stocks representing the temperatures of the liquids change depending on the rate of the temperature increase or decrease. Each rate equation is of the form coefficient*temp difference/mass.

A trial and error method can be used to find the value of the coefficient. We know that the black coffee cools to 52 degrees in 30 minutes. Trying values of 1 and 10 for the coefficient, we find that 1 results in too little cooling and 10 results in too much. It doesn't take long to settle on the value of 6.5.

Coffee Cooling Model Results
MinuteCoffeeCreamCreamed CoffeeFinal Mix

These results indicate that the final mix is about 3 degrees cooler than the creamed coffee. This conclusion was verified by advanced chemistry students who used the equation T-Ts = (To - Ts)*exp(-kt), where Ts is the temperature of the surroundings, To is the initial temperature of the substance, and T is the temperature of the substance at time t. The equation was used three times, once to find k given the facts for black coffee, and then, after adjusting this factor for the different masses, to find the final temperature for the cream and the creamed coffee. The value of k in this equation includes mass and, therefore, is not the same as the cooling coefficient in the model.

This problem exemplifies the three approaches to solving a scientific problem - experimental, computational, and theoretical. Providing students the opportunity to practice all three methods expands their tool kit for future problems.

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