Milk+and+Coffee+group+4

=Milk and Coffee Problem=

Group 4

Imagine that you have just poured yourself a mug of coffee (250mL) and since you prefer your coffee to be white, are about to add the milk (30mL), when the phone rings and you have to leave for 5 minutes. The question you ask yourself is if you want to drink the coffee as hot as possible, is it better to add the milk before you leave or after you return? What do you think? Does it make a difference?

Here is some information that will help you explore this further... The coffee when poured is 70 degrees Celsius. The milk is 5 degrees Celsius. The room stays at a constant temperature of 20 degrees Celsius. If the milk and coffee are left to stand separately, during the 5 minutes the coffee would have fallen to 60 degrees Celsius, the milk would have risen to 8 degrees Celsius.

Solution
In this problem I am going to assume that the rate of cooling/heating depends on the difference between the temperature of the substance and the ambient temperature, therefor it is going to be a changing rate. I am not going to assume that the milk and the coffee will behave the same way when it changes temperatures, so the cooling and heating rate of the milk and coffee are going to be investigated independently.

If you add the milk when you return...
(Vol of coffee/Vol of total)(temp of coffee) + (Vol of milk/Vol of total)(temp of milk) = temp total

(250/280)60 + (30/280)8 = __**54.4c**__

If you add the milk before you leave...
First we have to find the function for temperature change.

Parent function of cooling: exponential
 * t_final = t_initial k^x**

the temperature are not their real temperatures, it is their temperatures relative to the room temperature. k is a constant, might be different for milk and coffee x is the number of minutes


 * we are going to calculate the k for coffee first**

coffee t_initial: 70-20=50c coffee t_final: 60-20=40c

40 = 50k^5 .8 = k^5 k=.956

the k for coffee is .956


 * then we are going to find the k for milk, since it might be different**

milk t_initial: 70-20=50c milk t_final: 60-20=40c

-12 = -15k^5 .8 = k^5 k=.956

the k for milk is also .956

overall equation for temperature change
 * t_final = t_initial (.956)^x**

using this equation, we can calculate how the mixture of milk and coffee will cool

(Vol of coffee/Vol of total)(temp of coffee) + (Vol of milk/Vol of total)(temp of milk) = temp total

(250/280)70 + (30/280)5 = 63.0c after mixing

63.0-20 = 43.0c 43.0(.956)^5 = 34.4 34.4+20 = __**54.4c**__

Conclusion
It does not matter when you add the milk, the final temperature would not change.

Discussion
A graph showing the change in temperature to the different liquids over time.

With t approaching infinity, the t_final will approach 0, meaning that the temperature of the coffee would approach the temperature of the air around it when you leave it for a very long time.

If the coffee was in a room and the room is a closed system, then the room would heat up as the coffee cools. A closed system is when not energy/heat escapes or enters. The coffee will release heat as it cools and the air in the room would absorb the heat, but since the room is so much bigger than the coffee, it would warm up just a tiny bit.

The heating and cooling rate of something should also be determined by the amount of surface area it has, but after find that the equation for the temperature change is the same in the coffee and in the milk even when they have different volumes, we concluded that the impact of the change in surface area is so small that it can be neglected. We could not have calculated the surface area even if we wanted to since we do not know the shape of the containers containing the milk and the coffee. The total surface area when the milk and the coffee are separated should be greater than that of when the milk and the coffee is mixed together, so we speculate that in reality the temperature of the coffee should be different in the two situation. It would be impossible to determine which one would be hotter because the milk will heat faster when it is separated but the coffee will also cool faster when separated, and we do not know the shape of the containers. We also don't know if the milk is in a sealed plastic cup or in a jug or in a carton, and we don't know if the coffee is in a heat insulating Styrofoam cup or in a coffee mug.

If this were a real situation I would recommend bringing the coffee so you can drink it while talking on the phone, if you want to drink it at hot as possible.

Links
[|Wikipedia Page on Heat Transfer] [|Wikipedia page explaining the properties of heat]