Posted by Kelsie on Sunday, October 12, 2008 at 6:16pm.
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.
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
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
Related Questions
chemistry - how many ml of water at 10 degrees celsius must be added to 75 ml of...
Chemistry - A 100g sample of water at 90 degrees celsius is added to a 100g ...
Chemistry - A 5.0-g Sample of KBr at 25.0 degrees celsius dissolves in 25.0 ...
Chemistry - A coffee-cup calorimeter initially contains 125 g water at 24.2 ...
Chemistry - A sample of benzoic acid contaminated with NaCl and sand is to be ...
Chemistry - 20.7mL of ethanol (density=0.789g/mL) initially at 8.2 degrees ...
Chemistry - When 22.7mL of o.500 M H2SO4 is added to 22.7mL of 1.00 M KOH in a ...
Chemistry - A 5.0-g Sample of KBr at 25.0 degrees celsius dissolves 25.0 mL of ...
Chemistry - A 5.0g KBr sample, at 25 degrees celsius, dissolved in 25 ml of ...
chemistry - You drop a hot copper penny (500 degrees celsius) that has a mass of...
For Further Reading
