Question

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"`.

Answer 1

`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!

Answer 2

`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")`