# An unknown volume of water at 18.2 degrees celsius is added to 24.4 mL of water at 35.0 degrees celsius. If the final temperature is 23.5 degrees celsius what was the unknown volume? (assume no heat is lost to surroundings and D of water is 1.00g/mL)

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1. Heat gained by water + heat lost by water = 0.
unknown mass H2O x specific heat x (Tf-Ti) + known mass water x specific heat x (Tf-Ti) = 0
Tf = final T
Ti = initial T.
You have only one unknown; i.e., unknown mass H2O. Post your work if you get stuck.

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2. x= unknown mass of H2O

x(4.184)(23.5-35) + 17.008(4.184)(23.5-18.2)=0

-48.116x + 377.156= 0
x = 7.84g

d=1.00g/mL x=unknown volume
1.00= 7.84/x
x=7.84mL

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3. x= unknown mass of H2O

x(4.184)(23.5-35) + 17.008(4.184)(23.5-18.2)=0
I don't think so. Tf = final T; Ti = initial T.
a. Isn't the initial T 18.2 and not 35 for the unknown amount of water?
b. Where did the 17.008 come from? That must be the mass of water but that was 24.4 mL so it has a mass of 24.4 mL x 1.00 g/mL = 24.4 grams.
c. And delta T for the known amount of water is 23.5 - 35.0 isn't it?

-48.116x + 377.156= 0
x = 7.84g

d=1.00g/mL x=unknown volume
1.00= 7.84/x
x=7.84mL

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4. V = unknown volume of H2O, which equals the mass because the density is 1.00

V(4.184)(23.5-18.2) + 24.4(4.184)(23.5-35) = 0

22.1752V = 1174.0304

V = 52.94mL

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