If a pork roast must absorb 1500 kJkJ to fully cook, and if only 14% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO2CO2 is emitted into the atmosphere during the grilling of the pork roast?

You need to know the amount of heat generated by the combustion reaction.

Assuming propane as fuel, you can use thiis data:

C3H8(g)+5O2(g)---3CO2(g)+4H2O(g) ΔH= -2217 KJ

So when 3 moles of CO2 is emmitted 2217 kJ of heat is produced.

The molar wegiht of CO2 is 12 g/mol + 2 * 16 g/mol = 44 g/mol.

Then 3 mol * 44 g / mol = 132 g of CO2 are produced with 2217 kJ of heat.

Now you have to calculate how much energy you need to produce if only 12% is abosrbed by the pork

Energy absorbed by the pork = 12% *  total energy =>

total energy = energy absorbed by the pork / 0.12 = 1700 kJ / 0.12 = 14,166.67 kJ.

Now, state the proportion:

132 g CO2 / 2217 kJ = x / 14,166.7 kJ =>

x = 14,166.67 * 132 / 2217 = 843.48 g CO2.

843 g of CO2

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error