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September 16, 2014

September 16, 2014

Posted by **Jamie** on Sunday, October 12, 2008 at 2:59am.

1.One piece of copper jewelry at 111Â°C has exactly twice the mass of another piece, which is at 36.0Â°C. Both pieces are placed inside a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter (c of copper = 0.387 J/gÂ·K)?

2.When 25.5 mL of 0.615 M H2SO4 is added to 25.5 mL of 1.23 M KOH in a coffee-cup calorimeter at 23.50Â°C, the temperature rises to 30.17Â°C. Calculate Î”H per mole of H2SO4 and Î”H per mole of KOH. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.)

- Chemistry -
**DrBob222**, Sunday, October 12, 2008 at 5:17pm1.

Just assume any mass (like 2 g) for the first piece of jewelry which makes the second piece 1 g (2grams)0.

Then mass1 x specific heat x (Tf-Ti) + mass2(1 gram) x specific heat x (Tf-Ti) = 0

for mass 1st piece, m = 2 and Ti = initial T = 111

for mass 2nd piece, m = 1 and Ti = 36. Solve for Tf = final T.

- Chemistry -
**DrBob222**, Sunday, October 12, 2008 at 5:28pm2.

Write the balanced equation.

How much heat is given off in the neutralization reaction? That is

mass x specific heat x delta T.

You know mass (25.5 mL + 25.5 mL = 51 mL = 51 grams since the problem tells you the density is 1.00). Specific heat pure water is 4.184 J/g*C and you know delta T.

Then q/mols H2SO4 will give you the delta H/mol H2sO4.

q/mols KOH will give you the delta H/mol KOH.

- Chemistry -
**Chris**, Monday, November 15, 2010 at 2:59pmDrBob can you explain it more in depth?

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