Posted by **y912f** on Monday, February 23, 2009 at 5:40pm.

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

- physics -
**Damon**, Monday, February 23, 2009 at 5:44pm
Too many numbers for me tonight. method is correct.

- physics -
**y912f**, Monday, February 23, 2009 at 6:01pm
what about the answer?

- physics -
**Damon**, Monday, February 23, 2009 at 6:12pm
Not willing to do arithmetic, have flu and bronchitis, mind hazy.

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**y912f**, Monday, February 23, 2009 at 6:35pm
fine, thats a good excuse. thanks anyway

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**Anne**, Tuesday, April 28, 2015 at 11:09pm
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 didnt practice for. Ok ONE question of the three we didnt learn. So.....I cant really help you. Sorry.

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