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.
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.
3 answers
3.37
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).
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).
1 answer