Question

A sample of 0.5662 g of carbon is burned in oxygen in a bomb calrimeter, producing carbon dioxide, Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74C to 27.93C. What is the heat capacity of the calorimeter and its contents?

Answer 1

`C_(p)= (15.82)/(m_(calorimeter))\\\\c_(p)=7.64 kJ/gK`

Step-by-step explanation:

First we set the equation for the combustion:

`C_((s))+O_(2)_((g)) \longrightarrow CO_(2)_((g))`

As the equation is already balanced, we just need to look at the coefficients, that are 1 for all the elements of the reaction, so 1 mole of carbon produces 1 mole of carbon dioxide. Now we need to find how many moles are 0.5662 g of carbon:

`M_(C)=12.01 g/mole\\n_(C)=m_(C)/M_(C)\\n_(C)=0.047mole`

Know we proceed to know how many moles of carbon dioxide were produced. As we know the reaction is 1:1, the moles of carbon dioxide are 0.047.

Now we need to know the mass of the carbon dioxide produced:

`M_{CO_(2)}=44.01g/mole\\m_{CO_(2)}=M_{CO_(2)}*n_{CO_(2)}\\m_{CO_(2)}=2.07g`

From literature we find that the enthalpy of formation of carbon dioxide is -393.5 kJ/mole. We need it in terms of mass, so we convert it with the molar mass as follows:

`-393.5kJ/mole * M_{CO_(2)}=-8.94kJ/g`

With this we can continue with the calculation given the equation Q=q*m

So Q=18.51 kJ

To calculate the heat capacity we use Q=mCpΔT, therefore:

`C_(p)=(Q)/(m*\Delta T) \\\\C_(p)=(18.51)/(m*1.19)=(15.82)/(m_(calorimeter))`

As we don't know the mass of the calorimeter, we just leave the specific heat capacity as follows:

`c_(p)=(q)/(\Delta T)\\\\q=8.94kJ\\\Delta T=1.19\\c_(p)=7.64 kJ/gK`