Sucrose is commonly known as table sugar. The formula is C12H22O11. You burn 7.29 grams of sucrose in a bomb calorimeter. The temperature of the calorimeter and the water in it rises from 23.24 degrees celsius to 29.77 degrees celsius. The bomb's heat capacity had been determined to be 6.35 kJ/degrees celsius. There is 1400 grams of water in the calorimeter. Find the heat of combustion of sucrose in both kJ/gram and kJ/mole.
q to heat water from 23.24 to 29.77 =
mass H2O x specific heat H2O x delta T = xx J. Change to kJ.
q2 to heat calorimeter = 6.35 x delta T = yy kJ.
Add xx kJ + yy kJ to obtain total heat produced by 7.02 g sucrose. That gives you zz kJ/7.02 g. Convert to kJ/g and to kJ/mol. Post your work if you need more help.
To find the heat of combustion of sucrose, we can use the equation:
q = m * C * ΔT
where:
q is the heat transferred in joules (J)
m is the mass of the substance in grams (g)
C is the heat capacity of the calorimeter in joules per degree Celsius (J/°C)
ΔT is the change in temperature in degrees Celsius (°C)
First, let's calculate the heat transferred in joules (q):
q = m * C * ΔT
q = 1400 g * 6.35 kJ/°C * (29.77°C - 23.24°C)
To convert the given heat capacity of the bomb calorimeter from kJ/°C to J/°C, we multiply it by 1000:
q = 1400 g * 6.35 kJ/°C * (29.77°C - 23.24°C) * 1000
Next, we need to convert the grams of sucrose to moles using the molar mass of sucrose (C12H22O11). The molar mass of sucrose can be calculated by summing the atomic masses of its constituent elements:
Molar mass of sucrose (C12H22O11) = (12 * 12.01 g/mol) + (22 * 1.008 g/mol) + (11 * 16.00 g/mol)
Now, we have the moles of sucrose:
moles of sucrose = 7.29 g / molar mass of sucrose
Finally, we can calculate the heat of combustion per mole and per gram of sucrose:
Heat of combustion per mole = q / moles of sucrose
Heat of combustion per gram = q / grams of sucrose
Substituting the calculated values, we can find the answers to both questions.