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.)
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.
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.
DrBob can you explain it more in depth?
1. To solve this problem, we can use the principle of energy conservation. The heat gained by the colder piece of copper will be equal to the heat lost by the hotter piece of copper.
Let's denote the mass of the first piece of copper as m1, the temperature of the first piece as T1, the mass of the second piece as m2, and the temperature of the second piece as T2.
We know that the specific heat capacity of copper (c) is 0.387 J/g·K.
The equation for heat transfer is given by:
Q = mcΔT
Where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The heat gained by the colder piece is m1 * c * (final temperature - T1), and the heat lost by the hotter piece is m2 * c * (T2 - final temperature). Since the heat gained equals the heat lost, we can write the equation as:
m1 * c * (final temperature - T1) = m2 * c * (T2 - final temperature)
Substituting the given values, we have:
m1 * 0.387 * (final temperature - 36.0) = 2 * m1 * 0.387 * (111 - final temperature)
Simplifying the equation by canceling out the common terms:
(final temperature - 36.0) = 2 * (111 - final temperature)
Expanding and rearranging the equation:
final temperature - 36.0 = 222 - 2 * final temperature
3 * final temperature = 222 + 36.0
final temperature = (222 + 36.0) / 3
final temperature = 86.0°C
Therefore, the final temperature inside the calorimeter is 86.0°C.
2. To solve this problem, we can use the equation for heat transfer similar to the previous problem.
The heat gained by the solution (q) can be calculated using the formula:
q = m * c * ΔT
Where q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
We are given the volume of the solutions (25.5 mL), and we know that the density and specific heat capacity of the solution are the same as for pure water.
The density of water is 1 g/mL, so the mass of the solution can be calculated as:
mass = volume * density
mass = 25.5 mL * 1 g/mL
mass = 25.5 g
Using the given concentrations, we can calculate the moles of H2SO4 and KOH:
moles of H2SO4 = concentration * volume
moles of H2SO4 = 0.615 M * 25.5 mL
moles of H2SO4 = 15.65 mmol
moles of KOH = concentration * volume
moles of KOH = 1.23 M * 25.5 mL
moles of KOH = 31.31 mmol
Now, using the heat transfer equation, we can calculate q for each solution:
q(H2SO4) = moles of H2SO4 * c * ΔT
q(H2SO4) = 15.65 mmol * 25.5 g/mol * (30.17 - 23.50) °C * 4.18 J/g·K
q(H2SO4) = 1.53 kJ
q(KOH) = moles of KOH * c * ΔT
q(KOH) = 31.31 mmol * 25.5 g/mol * (30.17 - 23.50) °C * 4.18 J/g·K
q(KOH) = 3.06 kJ
The enthalpy change (ΔH) can be calculated by dividing the heat gained (q) by the number of moles:
ΔH per mole of H2SO4 = q(H2SO4) / moles of H2SO4
ΔH per mole of H2SO4 = 1.53 kJ / 15.65 mmol
ΔH per mole of H2SO4 = 97.8 kJ/mol
ΔH per mole of KOH = q(KOH) / moles of KOH
ΔH per mole of KOH = 3.06 kJ / 31.31 mmol
ΔH per mole of KOH = 97.7 kJ/mol
Therefore, the ΔH per mole of H2SO4 is 97.8 kJ/mol, and the ΔH per mole of KOH is 97.7 kJ/mol.