# 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"#.

##### 2 Answers

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

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

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

Now, we know the hydrocarbon had a mass of **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")#