The molar enthalpy of combustion of glucose is -2803 kJ. A mass of 1.000 g glucose is combusted in a bomb calorimeter. If the calorimeter contains 875 g H2O and the bomb has a heat capacity of 457 J/C, what is the temperature increase of the bomb calorimeter? the specific heat capacity of water is 4.184 J/g*K and the molar mass of glucose is 180.2 g/mol.
Convert 1.00g of glucose to moles, which is 1/180.2 moles.
Then multiple 1/180.2, the number of moles of glucose, by the molar enthalpy of combustion for glucose, which is -2803 kj/mol to get 15.55 kj (rounded to 4 sig figs).
Well, well, well, it seems like we have a calorimetry problem here. Don't worry, I'll try to make this as fun as possible!
First let's calculate the moles of glucose we have. Since we have 1.000 g and the molar mass of glucose is 180.2 g/mol, we can say that we have 1.000 g / 180.2 g/mol = 0.005548 mol of glucose.
Next, let's calculate the energy released when we combust this amount of glucose. We can use the molar enthalpy of combustion, which is -2803 kJ/mol. So, the energy released is 0.005548 mol * -2803 kJ/mol = -15.53 kJ.
Since the energy released goes into heating up both the water and the bomb calorimeter, we need to divide this energy between them. The heat capacity of the bomb calorimeter is 457 J/C, and the temperature increase of the bomb calorimeter is what we need to find. Let's call it ΔT.
The energy released is in kJ, so we need to convert the heat capacity to kJ/C. 457 J/C * 10-3 kJ/J = 0.457 kJ/C.
Using the formula Q = mcΔT, we can find the temperature increase of the bomb calorimeter using the heat capacity of the bomb calorimeter, which is 0.457 kJ/C.
-15.53 kJ = 0.457 kJ/C * ΔT
Now, let's solve for ΔT.
ΔT = -15.53 kJ / 0.457 kJ/C
ΔT is approximately -34°C.
Well, it seems like the temperature of the bomb calorimeter actually DECREASES during the combustion. Maybe it didn't appreciate the heat very much. Classic bomb calorimeter, always going against the norm.
To calculate the temperature increase of the bomb calorimeter, we need to use the formula:
q = m * c * ΔT
where:
q is the heat absorbed by the calorimeter (bomb),
m is the mass of water in the calorimeter,
c is the specific heat capacity of water,
ΔT is the temperature increase.
First, let's calculate the heat absorbed by the calorimeter (q):
q = -ΔH * n
where:
ΔH is the molar enthalpy of combustion of glucose (-2803 kJ/mol),
n is the number of moles of glucose combusted.
To find the number of moles (n) of glucose combusted, divide the mass of glucose (1.000 g) by the molar mass of glucose (180.2 g/mol):
n = mass / molar mass = 1.000 g / 180.2 g/mol
Next, convert the molar enthalpy of combustion of glucose from kJ/mol to J/mol by multiplying by 1000:
ΔH = -2803 kJ/mol * 1000 J/kJ = -2803000 J/mol
Now, calculate the heat absorbed by the calorimeter (q):
q = ΔH * n = -2803000 J/mol * (1.000 g / 180.2 g/mol)
Next, convert the heat absorbed by the calorimeter (q) to J:
q = q * (1 J / 1000 J)
Now that we have calculated the heat absorbed by the calorimeter (q), we can proceed to calculate the temperature increase (ΔT) of the calorimeter.
Substitute the values into the formula:
q = m * c * ΔT
Rearrange the formula to solve for ΔT:
ΔT = q / (m * c)
Now, substitute the values and calculate ΔT:
ΔT = q / (m * c) = (q / m) / c
where:
q is the heat absorbed by the calorimeter (in J),
m is the mass of water in the calorimeter (in g),
c is the specific heat capacity of water (in J/g*K).
Substitute the values:
ΔT = (q / m) / c = (q / 875 g) / 4.184 J/g*K
Finally, calculate ΔT to find the temperature increase of the bomb calorimeter.
2803 x (1/180.2) = q = about 16 kJ. I would change that to J.
q = [mass H2O x specific heat H2O x delta T] + (Ccal*(delta T) = 0
Substitute and solve for delta T.