Asked by y912f

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)(tf-ti)=M2(ti-tf)
10(tf-10)=160(90-tf)
10tf-100=160(90-160tf)
10t=160(90)+100-160tf
tf is the final eqquilibirum temperature
tf=85.29degreesC
Answered by Damon
Too many numbers for me tonight. method is correct.
Answered by y912f
what about the answer?
Answered by Damon
Not willing to do arithmetic, have flu and bronchitis, mind hazy.
Answered by y912f
fine, that's a good excuse. thanks anyway
Answered by Anne
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.
Answered by Michela
Anne did you really say you can't help someone 6 years later?
Answered by Kat
Michela, did you really awnser someone 3 years later?
Answered by gary
Kat, did you really just respond to someone 10 days later?
Answered by Kim
Gary did you really just reply to someone today?
Answered by Sultan
Kim did you really just reply to someone on the same day?
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