1.
mass H2O = 85 g + 85 g = ?
(mass H2O x specific heat H2O x (Tfinal-Tinitial) + Ccal*(Tfinal-Tinitial) = 0
Substitute and solve for Tf.
1) A quantity of 85 mL of .900 M HCl is mixed with 85 mL of .900 M KOH in a constant-pressure calorimeter that has a heat capacity of 325 J/C. If the initial temperatures of both solutions are the same at 18.24 degrees C, what is the final temperature of the mixed solution?
2) A 2.10-mole sample of crystalline acetic acid, initially at 17 degrees C, is allowed to melt at 17 degrees C and is then heated to 118.1 degrees C (its normal boiling point) at 1 atm. The sample is allowed to vaporize at 118.1 degrees C and is then rapidly quenched to 17 degrees C, so that it re-crystalizes. Calculate delta(H) for the total process described
3) The combustion of what volume of ethane (C2H6), measured at 23 degrees C and 752 mmHg, would be required to heat 855 g of water from 25 degrees C to 98 degrees C?
3 answers
3.
How much heat (q) is needed to heat the water.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) = approx 250,000 but that is an estimate as are all of the other numbers that follow.
Look up the heat combustion for ethane, probably given in kJ/mol
Convert that to the number of mols needed to provide the q from the first part, then convert mols to volume in L at the condition listed.
Post your work if you get stuck.
How much heat (q) is needed to heat the water.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) = approx 250,000 but that is an estimate as are all of the other numbers that follow.
Look up the heat combustion for ethane, probably given in kJ/mol
Convert that to the number of mols needed to provide the q from the first part, then convert mols to volume in L at the condition listed.
Post your work if you get stuck.
2. I've been thinking about this and I can't see a good reason why it isn't zero. It absorbs heat on the way up to vapor and releases heat on the way back down to the crystalline state. Think about that, is that right? If that is true it sure would save a lot of calculating time.