[hokychik]

"A chemical engineer studying the properties of fuels placed 1.150g of a hydrocarbon in a bomb of a calorimeter and filled it with O₂ gas. The bomb was immersed in 2.550L 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 (q_v) per gram of the fuel?"

I'm stuck on the last part of the problem. I used the -q_sys=q_surr and got q_surr=[m_h2o*c_h2o*(T_f-T_i)]+[C_cal*(T_f-T_i)] and I converted the volume of water to grams, and know the temperature change in Celsius is the same amount of change in Kelvin. For that part I got 39306.31 J. So I know that q_sys=-39306.31 J and also know that is equal to [m_fuel*c_fuel*(T_f-T_i)] but I don't know how to get that from the given information. Would someone please help me with this last bit?

[UG]

So I know that q_sys=-39306.31 J and also know that is equal to [m_fuel*c_fuel*(T_f-T_i)]

I don't think this is correct, there is no more of the fuel left. The combustion 0f 1.150 g of the hydrocarbon produced 39306 J of energy, all you need is to work out how much energy is produced when you burn 1.00 g of the fuel.

[hokychik]

To do that I would just do 39306.31J/1.150g. Maybe I messed up on the math somewhere. It asks for the answer in J/g (which is what that is) and in scientific notation, so i got 3.148*10⁴J/g

[UG]

so i got 3.148*10⁴J/g

And you got it wrong? Did you add a negative sign? What value of the heat capacity of water did you use?

[hokychik]

ahh I forgot the negative sign  thanks!