JAMB - Chemistry (2000 - No. 27)
0.46g of ethanol when burned raised the temperature of 50g of water by 14.3K. Calculate the heat of combustion of ethanol.
[C = 12, O = 16, H = 1, Specific heat capacity of water = 4.2 jg-1K-1]
[C = 12, O = 16, H = 1, Specific heat capacity of water = 4.2 jg-1K-1]
+3000 KJ mol-1
+300 KJ mol-1
-300 KJ mol-1
-3000 KJ mol-1
Explanation
Heat absorbed by water = (mCΔt) = (50 x 4.2 x 14.3) = 3003.0J = 3.003kJ
no of moles of ethanol = \(\frac{mass}{molar mass}\) = \(\frac{0.46}{46}\) = 0.01 mol.
Therefore, heat of combustion of ethanol = \(\frac{heat absorbed by water}{no of mole of ethanol}\) = \(\frac{-3.003}{0.01}\) = - 300.3kj/mol
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