when 1.00g of sucrose C12H22O11 was combusted in a calorimeter to give CO1 and H20, the temperature rose from 25.00 to 27.32 C. The calorimeter itself had a heat capacity of 837JC and was submersed in 1500.0g of water. Find the heat of combustion of sucrose.
How would I go about finding the answer?
q = [mass H2O x specific heat H2O x (delta T)]+[Ccal*delta T] = 0
q=(1500.0g H2O)(4.184J K^-1)(2.32K)
To find the heat of combustion of sucrose, you can use the formula:
q = mcΔT
where:
q = heat energy (in joules)
m = mass (in grams)
c = specific heat capacity (in joules per gram-degree Celsius)
ΔT = change in temperature (in degrees Celsius)
In this case, the heat energy is generated by the combustion of sucrose and is absorbed by both the calorimeter and the water. The calorimeter has a heat capacity, which is the amount of heat required to raise its temperature by 1 degree Celsius. The water has a specific heat capacity, which is the amount of heat required to raise the temperature of 1 gram of water by 1 degree Celsius.
First, we need to calculate the heat absorbed by the calorimeter. To do this, we multiply the heat capacity of the calorimeter (837 J/C) by the change in temperature (27.32 - 25.00 = 2.32 C).
q_calorimeter = (837 J/C) * (2.32 C)
Next, we need to calculate the heat absorbed by the water. To do this, we multiply the mass of the water (1500.0 g) by its specific heat capacity (4.18 J/gC) and the change in temperature (27.32 - 25.00 = 2.32 C).
q_water = (1500.0 g) * (4.18 J/gC) * (2.32 C)
Finally, to find the heat of combustion of sucrose, we subtract the heat absorbed by the calorimeter and the heat absorbed by the water from the total heat energy generated during the combustion. The combustion of 1.00 g of sucrose releases a certain amount of heat, which we need to find.
Heat of combustion of sucrose = q_combustion = q_total - (q_calorimeter + q_water)
Therefore, to find the heat of combustion of sucrose, you need to calculate q_calorimeter, q_water, and q_combustion using the formulas and values provided.