A chemical engineer studying the properties of fuels placed 1.290 g of a hydroca

Question

A chemical engineer studying the properties of fuels placed 1.290 g of a hydrocarbon in the bomb of a calorimeter and filled it with 02 gas. The bomb was immersed in 2.550 L of water and the reaction initiated. The water temperature rose from 20.00 degree C to 23.55 degree C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat of reaction for combustion (q y) per gram of the fuel? (d for water = 1.00 g/mL; c for water = 4.1S4 J/g- degree Enter your answer in scientific notation.

Explanation / Answer

chemical reaction :

hydrocarbon + O2 ====> product + heat

heat generated = heat transferred to water and to calorimeter

mass of water = volume * density = 2550 ml * 1 gm / ml = 2500 gm ......( as 1ml = 1cm3 )

heat gained by water = mass * specific heat * temperature difference

                                = 2500 g x 4.184 x (23.55-20.00) = 37133 J

heat gained by calorimeter = 403 x (23.55-20.00) = 1430.65 J

total heat from the reaction = 37133+1430.65 = 38563.65 = 38564 J

The mass of HC is 1.29 g

so the heat of combustion = -38564 J / 1.29 g = -29894.58 J/g = -2.99 *104 J /g

negative sign is because the energy is given off.

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