When a 0.650 g sample of olive oil is burned in a calorimeter, the heat released increases the temperature of 370 g of water from 22.7 ∘C to 38.9 ∘C.

Question

When a 0.650 gg sample of olive oil is burned in a calorimeter, the heat released increases the temperature of 370 gg of water from 22.7 ∘C∘C to 38.9 ∘C∘C.

Part A

What is the energy value of the olive oil in kJ/gkJ/g?

Answer 1

q = mC∆T

q = heat = ?

m = mass = 370 g H2O

C = specific heat = 4.184 J/gº

∆T = change in temperature = 38.9 - 22.7 = 16.2º

Solving for q (heat), we have ...

q = (370 g)(4.184 J/gº)(16.2º) = 25,079 J

The question asks for q in terms of kJ/g, so we will convert this answer to kJ and divide by the 0.65 g ...

25,079 J x 1 kJ / 1000 J = 25.079 kJ

25.079 kJ / 0.650 g = 38.6 kJ / g (to 3 sig. figs.)

Answer 2

Total energy = specific heat of water * mass of water * temperature difference =

4.184 kJ/(kg*∘C) * (370 g * 0.001 kg/g) * (38.9 ∘C - 22.7 ∘C) =

(4.184 * 0.370 * 16.2) kJ = 25.078896 kJ