001 (part 1 of 2) 10.0 points
A 1.37-gram sample of compound X (of MW 88.0 g/mol) was burned in a bomb calorime- ter containing 1700 g of water. A temperature rise of 0.95◦C was observed. ∆Erxn for this re- action is −475 kJ/mol of X. What is the heat capacity of the calorimeter hardware (not the water)?
Answer in units of J/◦C
PART 2
What is the water equivalent of the calorime- ter hardware in the previous question?
Answer in units of grams
475,000*(1.36/88.0) = [mass H2O x specific heat H2O x delta T] + [Ccal*0.95]
Substitute and solve for Ccal.
Water equivalent of calorimeter is mass H2O that equals heat capacity of the calorimeter.
Ccal = mass H2O x specific heat H2O.
Solve for mass H2O.
To determine the heat capacity of the calorimeter hardware, you can use the formula:
q = m * c * ΔT
where:
q is the heat absorbed or released by the system
m is the mass of the water in the calorimeter
c is the specific heat capacity of water
ΔT is the temperature change observed
In this case, the heat absorbed by the water can be calculated using:
q_water = m_water * c_water * ΔT
Given:
m_water = 1700 g
c_water = 4.18 J/(g·°C)
ΔT = 0.95°C
Substituting the values into the equation:
q_water = (1700 g) * (4.18 J/(g·°C)) * (0.95°C)
Calculating q_water:
q_water = 6751.1 J
Since the heat absorbed by the water is equal to the heat released by the compound X, we can also write:
q_water = q_compound_X
Using the given ∆Erxn for this reaction is −475 kJ/mol of X, we can convert it to J/mol:
∆Erxn = -475 kJ/mol * (1000 J/1 kJ) = -475,000 J/mol
To find the number of moles of X that was burned, we can use the molar mass of X and the mass given:
moles_X = (1.37 g) / (88.0 g/mol)
Calculating moles_X:
moles_X = 0.01557 mol
Now, we can calculate the heat released by the compound X:
q_compound_X = ∆Erxn * moles_X
Substituting the values in:
q_compound_X = (-475,000 J/mol) * (0.01557 mol)
Calculating q_compound_X:
q_compound_X = -7,399.35 J
Since q_water = q_compound_X, we can conclude that:
6751.1 J = -7,399.35 J + q_calorimeter_hardware
Now, to find the heat capacity of the calorimeter hardware, we rearrange the equation:
q_calorimeter_hardware = 6751.1 J - (-7,399.35 J)
q_calorimeter_hardware = 14,150.45 J
Therefore, the heat capacity of the calorimeter hardware is 14,150.45 J/°C.
PART 2:
The water equivalent of the calorimeter hardware is defined as the mass of water that would absorb the same amount of heat as the calorimeter hardware during a temperature change.
The formula to calculate water equivalent is:
Water Equivalent = heat capacity of the calorimeter hardware / specific heat capacity of water
Given:
Heat capacity of the calorimeter hardware = 14,150.45 J/°C
Specific heat capacity of water = 4.18 J/(g·°C)
Substituting the values into the formula:
Water equivalent = 14,150.45 J/°C / 4.18 J/(g·°C)
Calculating the value of water equivalent:
Water equivalent = 3383.97 g
Therefore, the water equivalent of the calorimeter hardware in the previous question is 3383.97 grams.
To find the heat capacity of the calorimeter hardware (not the water), we can use the equation:
∆Erxn = qcal + qwater
Where:
∆Erxn = the change in energy of the reaction (in this case, -475 kJ/mol of X)
qcal = the heat absorbed or released by the calorimeter hardware
qwater = the heat absorbed or released by the water
Since we want to find the heat capacity of the calorimeter hardware, we need to isolate qcal in the equation:
qcal = ∆Erxn - qwater
To find qwater, we can use the equation:
qwater = mass of water * specific heat of water * temperature change
Given information:
Mass of water = 1700 g
Specific heat of water = 4.18 J/g°C (assuming it's at room temperature)
Temperature change = 0.95°C
Plugging in the values:
qwater = 1700 g * 4.18 J/g°C * 0.95°C
qwater = 6441.1 J
Now, substitute the value of qwater into the previous equation to find qcal:
qcal = -475 kJ/mol - 6441.1 J
To convert kJ to J, we multiply by 1000:
qcal = (-475 kJ/mol * 1000) - 6441.1 J
qcal = -475000 J/mol - 6441.1 J
To convert from J/mol to J/°C, we need to divide the result by the molar mass of compound X:
qcal = (-475000 J/mol - 6441.1 J) / (88.0 g/mol)
qcal ≈ -5501 J/°C
So, the heat capacity of the calorimeter hardware (not the water) is approximately -5501 J/°C.
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To find the water equivalent of the calorimeter hardware, we need to understand that the water equivalent is the mass of water that has the same heat capacity as the calorimeter hardware.
The equation to find the water equivalent is:
Water equivalent = (qcal) / (specific heat of water * temperature change)
Plugging in the known values:
qcal = -5501 J/°C
Specific heat of water = 4.18 J/g°C (assuming it's at room temperature)
Temperature change = 0.95°C
Water equivalent = (-5501 J/°C) / (4.18 J/g°C * 0.95°C)
Water equivalent ≈ -1426 g
Therefore, the water equivalent of the calorimeter hardware in the previous question is approximately -1426 grams.