In these problems one material loses heat and the other gains heat. In this case, the warm coffee loses heat and the cold milk gains heat. All of that exchange must be zero.
heat lost by coffee + heat gained by milk = 0
heat lost by coffee is
mass coffee x specific heat coffee x (Tfinal-Tintial)
heat gained by milk is
mass milk x specific heat milk x (TfinalTinitial)_
You put those together like this
[mass coffee x sp.h. coffee x (Tf-Ti)] + [mass milk x sp.h. milk x (Tf-Ti) = 0
Now you substitute the numbers into the above. We are assuming milk/coffee are really water with sp.h of each = 4.184 J/g*C. And since the density of water is 1.0 g/mL then the 250 mL coffee will be 250 g and the mL milk will be that many grams milk.
[250g coffee x 4.184 J/g*C x (Tf-Ti)] + [mass milk x 4.184 J/g*C x (Tf-Ti)] = 0
Ti for coffee is 95 C
Tf for coffee is 90 C
Ti for milk is 10 C
Tf for milk is 90 C (Tfinal for milk must be the same as Tfinal for coffee since they are mixed.)
The only unknown in that equation is mass milk (in grams) and with the density being 1.0 g/mL that will be the number of mL. If you post your work I'll be happy to check you answer.
PLEASE help because I'm really confused on how to do the entire thing. I don't know how to find the mL. Am I suppose to do anything with the 250mL?
A person stops in every day and orders 250 mL of house coffee at precisely 95°C. but enough milk at 10°C to drop the temperature of the coffee to exactly 90° C.
Calculate the amount of milk (in mL) the we must add to reach this temperature. Assume that the coffee and milk have the same specific heat capacity: 4.184 J/(g × °C). Assume that they also have the same density: 1.0 g/mL.
1 answer