1.
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.
Need Help with these two problems
1.One piece of copper jewelry at 111°C has exactly twice the mass of another piece, which is at 36.0°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/g·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 23.50°C, the temperature rises to 30.17°C. Calculate ΔH per mole of H2SO4 and ΔH 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.)
3 answers
2.
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.
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.
DrBob can you explain it more in depth?