# 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".

Jul 15, 2017

$q = - {C}_{c a l} \cdot \Delta T$

q = (403J)/(°C)*3.55°C

${q}_{s o \ln} = - {q}_{r x n}$

${q}_{r x n} = \frac{- 1.43 k J}{1.320 g}$

= (-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!

Jul 16, 2017

$\Delta {E}_{r x n} = {q}_{V , r x n} = - \text{29.78 kJ/g}$

The idea here is that a bomb calorimeter is one that is of a constant volume, with associated heat flow ${q}_{V} = \Delta E$, 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:

${q}_{V , c a l} = {C}_{V , c a l} \Delta T$,

where ${C}_{V , c a l} = \text{403 J/K}$ is the constant-volume heat capacity for JUST the calorimeter by itself and $\Delta T$ is the change in temperature of the system.

or...

${q}_{V , w} = {m}_{w} {\overline{C}}_{V , w} \Delta T$,

where ${m}_{w}$ is the mass of the water, ${\overline{C}}_{V , w} = \text{4.184 J/g"^@ "C}$ is the specific heat capacity of the water, and $\Delta T$ for the water is assumed to be the same as for the bomb.

We assume the density of water to be $\text{1000. g/L}$ for simplicity so that its mass is $\text{2550 g}$.

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

${q}_{V} = {q}_{V , c a l} + {q}_{V , w}$

$= {C}_{V , c a l} \Delta T + {m}_{w} {\overline{C}}_{V , w} \Delta T$

$= \left({C}_{V , c a l} + {m}_{w} {\overline{C}}_{V , w}\right) \Delta T$

$= \left[\text{403 J/"^@ "C" + ("2550 g")("4.184 J/g"^@ "C")] (23.55^@ "C" - 20.00^@ "C}\right)$

$=$ $\text{39306.3 J}$

And thus, the heat of reaction at constant volume is $- \text{39306.3 J}$.

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

$\textcolor{b l u e}{\Delta {E}_{r x n} = {q}_{V , r x n}} = - \text{39306.3 J"/"1.320 g}$

$=$ $- \text{29777.5 J/g}$

$=$ $\textcolor{b l u e}{- \text{29.78 kJ/g}}$