What is the change in internal energy for the combustion of #"1.320 g"# of a hydrocarbon in a bomb surrounded by #"2.55 L"# of water if the temperature rises from #20.00^@ "C"# to #23.55^@ "C"#? Assume the heat capacity for the dry device is #"403 J/K"#.

#q = -C_(cal)*DeltaT#

#q = (403J)/(°C)*3.55°C#

#q_(soln) = -q_(rxn)#

#q_(rxn) = (-1.43kJ)/(1.320g)#

#= (-1084J)/g*(eV)/(1.602*10-19J) = 6.767*10^21 eV#

Maybe your professor wanted you to relate the water in the calorimeter to this somehow, but that's how I know to find it. Someone correct me if you know how!

#DeltaE_(rxn) = q_(V,rxn) = -"29.78 kJ/g"#

The idea here is that a bomb calorimeter is one that is of a constant volume, with associated heat flow #q_V = DeltaE#, the change in internal energy #E#.

We're given the volume of water for a reason; we have to treat

• the heat flow out from the reaction (system) into the bomb (surroundings), and
• the heat flow out from the reaction (system) into the water (surroundings),

separately.

Recall that the heat flow is given by:

#q_(V,cal) = C_(V,cal)DeltaT#,

where #C_(V,cal) = "403 J/K"# is the constant-volume heat capacity for JUST the calorimeter by itself and #DeltaT# is the change in temperature of the system.

or...

#q_(V,w) = m_wbarC_(V,w)DeltaT#,

where #m_w# is the mass of the water, #barC_(V,w) = "4.184 J/g"^@ "C"# is the specific heat capacity of the water, and #DeltaT# for the water is assumed to be the same as for the bomb.

We assume the density of water to be #"1000. g/L"# for simplicity so that its mass is #"2550 g"#.

Thus, the total heat flow from the reaction (with respect to its surroundings) is given by:

#q_V = q_(V,cal) + q_(V,w)#

#= C_(V,cal)DeltaT + m_wbarC_(V,w)DeltaT#

#= (C_(V,cal) + m_wbarC_(V,w))DeltaT#

#= ["403 J/"^@ "C" + ("2550 g")("4.184 J/g"^@ "C")] (23.55^@ "C" - 20.00^@ "C")#

#=# #"39306.3 J"#

And thus, the heat of reaction at constant volume is #-"39306.3 J"#.

Now, we know the hydrocarbon had a mass of #"1.320 g"#, so the heat of reaction at constant volume (or internal energy of reaction) per gram of hydrocarbon is just:

#color(blue)(DeltaE_(rxn) = q_(V,rxn)) = -"39306.3 J"/"1.320 g"#

#=# #-"29777.5 J/g"#

#=# #color(blue)(-"29.78 kJ/g")#