Energetics Question (Chemistry)?

The solution is as follows :

Mass of water
= (500 cm³) × (1.00 g/cm³)
= 500 g

Energy value per 1.13 g of the granola bar
= Energy absorbed by the water
= m c ΔT
= (500 g) × (4.18 J/g °C) × [(28.0 - 18.5) °C]
= 19855 J
= 19855/1000 kJ

Energy value per g of the granola bar
= (19855/1000 kJ) × (1/1.13)

Energy value per 100 g of the granola bar
= (19855/1000 kJ) × (1/1.13) × 100
= (19855/1000 kJ) × (100/1.13) ………. That is you asked how this working was done
= 1760 kJ (to 3 sig. fig.)

Hence, energy value = 1760 kJ/100 g

make sure you include your units in your calcs.

we start with this equation
.. Q = m * Cp * dT

and we apply that to just the water
.. Q = (500cm³ x 1g/cm³) * (4.184 J/g°C) * (28.0°C - 18.5°C) * (1kJ / 1000J) = 19.87 kJ

and that is for 1.13g so

.. 19.87 kJ.. .100... (19.87 * 100 / 1.13) kJ.... .1758 kJ
. ---- ---- --- x ----- = ----- ----- ---- ---- ---- --- = ---- ---- ----
.. .1.13g.. .. ..100... . ... ... . 100g... ... .... ... .. .. 100g

note..
.. I multiply by 100 / 100.. ..which = 1.. right? so I can do that
.. I leave one of the "100's" in the denominator.. I don't mess with it!
.. the numerator is 19.87 kJ * 100 / 1.13 = 1758 kJ.. again, the 100g stays in the denominator


*******
as to sig figs,
.. 500 in 500cm³ of H2O has 1.... <==== are you sure it's not "500.cm³".. with a decimal pt?
.. 1.13 in 1.13g has 3 sig figs
.. 28.0°C - 18.5°C = 9.5°C with 2 sig figs

so IF that 500 cm³ has a decimal point, then the answer is limited to 2 sig figs from the temp
.. final answer = 1.8x10³ kJ/100g

but IF that 500 does not a decimal pt, then the answer is limited to 1 sig fig
.. final answer = 2x10³ kJ/100g
..