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In an experiment, a student heated 550 g of water from 25oC to 91oC using 0.13 g of glycerol monostearate, an emulsifying agent used in the production of margarine.  If it is assumed that all the heat energy was absorbed by the calorimeter water, calculate the experimental enthalpy of combustion for glycerol monostearate in kJ/g.   

5 years ago

Answered By Leonardo F

We need to apply the concept of heat gained by the water = heat released by the combustion of the glycerol. Given that the heat capacity of the water is approximately 4.19 kJ/kg.K, we can calculate the heat absorbed (Q) by the water:

Q = (m)(Cp)(T2 - T1) where:

m is the mass of the water; Cp is the heat capacity of the water; T2 is the final temperature of the water and T1 is the initial temperature. The mass of the water is 550 g or 0.55 kg:

Q = (0.55)(4.19)(91-25)

Q = 152.1 kJ

Since the mass used of glycerol is 0.13 g, we need to divide the heat obtained in the last step by this value:

Q = 152.1 / 0.13 = 1.2×10^3 kJ/g

Hence, the experimental enthalpy of combustion of this chemical compound is approximately 1200 kJ/g.


5 years ago

Answered By Leonardo F

Just one last detail: the enthalpy of combustion of the glycerol is negative, because it releases energy (exothermic)