What is the final temperature reached by the addition of 11.4 grams of steam at 100C to 681 grams water at 25C in an aluminum calorimeter having a mass of 781 grams?

Question

C of water=1 cal/gc
C of aluminum= 0.22 cal/gc
L steam= 540cal/g

My Attempt

Qloss=Qgain
Qst+Qhw=Qcup+Qcw
ml+mcT=mcT+mcT

after plugging the numbers I'm not sure. All three final temps are the same at equilibrium and the cup and cold water start at 20. The hot water on the steam side starts at 100c. How do I solve? Thank you very much!!

Answer 1

I don't see how you handled the heat of vaporization.

Sum of heats gained=0
11.4(-Hv)+681cwater*(Tf-25)+781cAl*(tf-100)=0

-11.4*540+681*1*(Tf-25)+781*.22*(Tf-100)=0

do the algebra, solve for Tf.

Answer 2

Thanks for answering so quickly. I know there are different ways to doing things and I'm by no means an expert but don't we also have to account for the water that was previously steam, in this case 11.4 grams of water?