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

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

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Answer 1 · 2017-07-15T23:41:30

$q = -C_(cal)*DeltaT$

$q = (403J)/(°C)*3.55°C$

$q_(soln) = -q_(rxn)$

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

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

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 · 2017-07-16T00:39:00

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

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 $E$ .

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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:

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

$=$ $-"29777.5 J/g"$

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

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