A chemical engineer studying the properties of fuels placed 1.850 g of a hydroca

Question

A chemical engineer studying the properties of fuels placed 1.850 g of a hydrocarbon in the bomb of a calorimeter and filled it with O2 gas. The bomb was immersed in 2.550 L of water and the reaction initiated. The water temperature rose from 20.00 C to 23.55 C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat of reaction for combustion per gram of fuel? (d for water = 1.00 g/mL; c for water = 4.184 J/gC.) Please try to provide answer in scientific notation (I think that's where my mix up lies.)

Explanation / Answer

heat required to raise temp of water to 23.55 from 20 = specific heat x temp change x mass = 4.18 x(23.55-20) x (2.55x1000) = 37839.45 Joules heat required for calorimeter raise in temp = 403 x(23.55-20) = 1430.65 Joules total heat = (37839.45+1430.65) = 39270.1 Joules = heat of combustion for 1.85 gm fuel heat of combustion per gm fuel = 39270.1/1.85 = 21227.081 Joules = 21.22708 KJ

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