Original Question: A 2.00-g sample of palm tree oil (a component of some biodiesel fuels) was placed in a sealed chamber within a calorimeter (Figure 1) containing 1.80 kg of water. The sample was ignited and allowed to react with oxygen in a combustion reaction. The heat released by the reaction caused the surrounding water to increase in temperature from 24.1∘C to 34.3∘C. Assume that no heat escaped from the calorimeter or was absorbed by the container walls.
Part B) Given that the specific heat of water is 4.184J/g∘C, how much heat (in kJ) was absorbed by the water?
Answer = 76.8 kJ which is correct
Part C) In this case, the chemical reaction can be defined as the system. What is the value for qsystem?
Answer = -76.8 kJ Which is correct
Part D) What is the experimental value of the heat of combustion of palm tree oil in units of kJ/g?
Can someone please explain how to set up this calcualtion. Thank you.
Mass of palm tree oil 2.0 g
Mass of water = 1.80 kg = 1800 g
Increase in temperature, T = 34.3 - 24.1
= 10.2 oC
Part B) Specific heat of water, Cp = 4.184J/g oC.
Heat absorbed by water, q = m*Cp*T
= 1800*4.184*10.2
= 76818.24 J
q = 76.8 kJ
Part C) Since, heat escaped from the calorimeter or was absorbed by the container walls.
q system + q water = 0
q system + 76.8 kJ = 0
q system = - 76.8 kJ
Part D) Total heat of the system, q system = - 76.8 kJ
Heat of combustion of palm tree oil = q system / mass of palm tree oil
= - 76.8 kJ / 2.00-g
Heat of combustion of palm tree oil = - 38.4 kJ/g
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