Question

The “steam” above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature of 250 g of hot coffee initially at $90.0^\circ\textrm{C}$ if 2.00 g evaporates from it? The coffee is in a Styrofoam cup, so other methods of heat transfer can be neglected.

Final Answer

$85.7^\circ\textrm{C}$

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Video Transcript

This is College Physics Answers with Shaun Dychko.
We have some coffee in an insulated cup which means the only way that heat is going to leave is through this phase change of some of the coffee that is evaporating. So, there is a total of 250 grams to start with and two grams of that will evaporate leaving 248 grams left behind which we call

*m*subscript*w*for water because the coffee is going to have the properties of water, the same specific heat of water 4186 joules per kilogram per Celsius degree and the latent heat of vaporization for this coffee that is evaporating is 2256 joules per kilogram and this number is not precisely true because that's true only when the water is at 100 degree Celsius but 90 degree Celsius is pretty close to that and so we will take that to be the latent heat of vaporization. So, the initial temperature is 90 degrees and we are asked to figure out what will the final temperature of the coffee be after 2 grams of it evaporates. So, there is going to be a amount of heat that is lost and that’s why the negative sign is there because it’s a loss in thermal energy equal to the mass of coffee remaining times it specific heat times the final temperature minus the initial temperature and that heat loss is going to be the heat gained by the vapour that is evaporating, so the mass that’s evaporating times the latent heat of vaporization and then we are gonna solve this for*T f*coz we want to know what the final temperature of the coffee is. So we multiply through by*m w**C w*and then we have positive*m w**C w**T initial*minus*m w**C w*times*T final*and then we will take this term to the left side and then this term to the right side and then switch the sides around so that we have the unknown terms in the left and we are dividing both sides by*m w**C w*and that isolated*T f*. So*T final*is mass of the water remaining times its specific heat times initial temperature minus the mass of the vapour that evaporates times the latent heat of vaporization divided by the mass of the water times specific heat of water. So that’s 248 times ten to the minus three kilograms so that’s 248 grams converted to kilograms times specific heat of water times initial temperature of 90 degrees Celsius minus the two grams that evaporates written in kilograms times 2256 times 10 to the 3 joules per kilogram divided by 248 grams times specific heat of water and this gives a final temperature of 85.7 degree Celsius.