physics
posted by y912f on .
please tell me if this is done correctly
Find the final equlibirum temperature when 10.0 g of milk at 10.0 degrees C is added to 1.60 * 10^2 g of coffee with a temperature of 90.0 degrees C. Assume the specific heats of coffee and milk are the same as for water(Cp,w=4.19J/g*degC), and disregard the heat capacity of the container.
m=10 gm
ti(milk)=10degC
M2=1.6*10^2
g=160g
ti(coffee)=90degC
Cp=4019J/gdegC
m c/p deltaT = M2c/p deltaT
(10)(tfti)=M2(titf)
10(tf10)=160(90tf)
10tf100=160(90160tf)
10t=160(90)+100160tf
tf is the final eqquilibirum temperature
tf=85.29degreesC

Too many numbers for me tonight. method is correct.

what about the answer?

Not willing to do arithmetic, have flu and bronchitis, mind hazy.

fine, that's a good excuse. thanks anyway

idk the answer i am trying to find the answer for my own homework. GUESS WHAT?
chicken butt. jk. I need help on my hw too since i got work we didn't practice for. Ok ONE question of the three we didn't learn. So.....I cant really help you. Sorry.