Question

a chemical engineer studying the properties of fuels placed 1.180 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 20.00C to 23.55C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat of the reaction for combustion 9qV) per gram of the fuel? d for water = 1.00 g/mL, c for water= 4.184 j/g*C) Enter your answer in scientific notation

Answer 1

Here ; the calorimeter also was initially at 20.00 C and increased in temperature to 23.55 C, the same as the water. So the temperature change

= 23.55-20.00 = 3.55 C or3.55 K

Heat of calorimeter = 403 J/K *dt

= 403 J/K * (3.55 K)

= 430.6 J (3.55 deg K)

= 1430.6 J

Mass of water = density * volume

= 1.00 g/mL *2.550 L *1000 mL/1.0 L

=2550 g

Heat of water=mCdT

=2550 g*4.184 J/g*C*3.55 C

=37875.66 J

The total energy = 1430.6 J+ 37875.66 J

= 39306.26 J.

= 39.31 KJ

Here 1.180 g of a hydrocarbon so to calculate the heat of the reaction for combustion per gram of the fuel divide the total heat by the amount of fuel:

=39306.26 J/ 1.180g

= 33310.38 J/ g

= 3.3*10^4 J/g