Question

# Original Question: A 2.00-g sample of palm tree oil (a component of some biodiesel fuels) was...

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/gC, 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