Sweaty Iced Coffee Experiment

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The ‘Sweaty’ Iced Drink
Physics Around Campus

Jack Turner

Physics 102
Spring 2016

If you have ever had a cold drink on a hot day you have probably seen, and felt, the phenomenon that is taking place in the picture above. It seems as though over time, the longer you let your drink sit out the more it “sweats” and the more water there is on the outside of the cup than there was before. It looks like the water practically comes through the cup, or out of thin air, and ends up on the outside of the cup until the drink reaches room temperature. This picture of my Dutch Bros Iced Coffee is exactly that same phenomenon of a “sweating” drink. This photo was taken on a somewhat humid day in mid-April in my apartment …show more content…

Q is negative in this case because the water going from vapor to liquid loses energy to its surroundings during the phase change. One issue of calculating the energy given off by the condensation reaction for the picture of my iced coffee is that we do not know the mass of the water vapor being converted into water droplets. To make up for this I looked up the average humidity for Spokane in the month of April and found the mass, in grams, of the water vapor per one pound of air based on the humidity and temperature of the air (engineering tool box). From this information we can then calculate the energy given off in our specific phase change of the water vapor onto the cup containing the iced coffee. If the temperature in the room were about 80 degrees Fahrenheit, or 27 degrees Celsius, the calculated mass of the water vapor per pound of air would be about 8.1 grams or 0.0081 kg. The latent heat for condensation is the same as the latent heat for vaporization and it is 22.6 x 10^5 J/kg. Using the equation for the amount of energy given off by a pound of air reacting with the cold surface of the cup causing the phase change is as …show more content…

This energy then can be used in the heat conduction equation to see how much the condensation of the water vapor around the cup of iced coffee changes the temperature of the cup. This equation is Q=(kA∆T)t/L where k is the thermal conductivity of the medium the heat passes through, in this case it is a plastic cup so about 0.3 [J/(s*m*degrees celcius)]. A is the surface area of the cup which in this case is about 1337.7 cm2. Time we can vary but for this example we will say 30 minutes or 1800 seconds. L is length and for the length of the plastic cup we just use the thin distance that he heat had to pass through of about 0.002 meters. Finally with Q as 18,306 Joules we can rearrange to solve for the change in temperature and see how the condensation of water on the surface of the cup effected the temperature of the cup. After computing this it is seen that the change in temperature of the cup was about 0.05 degrees Celsius or about one degree

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