In a bomb calorimeter experiment, the heat capacity can be determined by analyzing the combustion of a known mass of a substance, in this case, ethane (C₂H₆). The heat of combustion for ethane is given as 1,060 kJ/mol, which is an exothermic reaction, indicating that the heat released will be negative when calculating the heat absorbed by the calorimeter.

To begin, we convert the heat of combustion from kilojoules to joules, knowing that 1 kJ equals 1,000 J. Thus, the heat of combustion becomes:

$-\mathrm{1,060,000}J/mol$.

Next, we need to calculate the number of moles of ethane burned. Given that the mass of ethane is 12.13 grams, we first find the molar mass of ethane:

For ethane, the molar mass is calculated as follows:

• Carbon: 2 × 12.01 g/mol = 24.02 g/mol
• Hydrogen: 6 × 1.008 g/mol = 6.048 g/mol

Adding these together gives:

$30.07g/mol$.

Now, we can calculate the number of moles of ethane:

$moles\; of\; ethane=12.1330.07=\; 0.4034\; \backslash text\{\; moles\}$.

Using the number of moles, we can calculate the heat absorbed by the calorimeter:

$q\; (calorimeter)=moles\times heat\; capacity\times \Delta T$.

Here, the specific heat capacity of the calorimeter is approximately 52.63 J/(mol·°C), and the temperature change (ΔT) is 15.2 °C. Thus, the heat absorbed by the calorimeter can be expressed as:

$q\; (calorimeter)=\; 0.4034\; \backslash text\{\; moles\}\; \backslash times\; 52.63\; \backslash text\{\; J/(mol·°C)\}\; \backslash times\; 15.2\; \backslash text\{\; °C\}$.

Calculating this gives:

$q\; (calorimeter)=\; 322.71\; \backslash text\{\; J\}$.

Now, applying the principle of conservation of energy, we set the heat released by the combustion equal to the heat absorbed by the calorimeter:

$q\; (combustion)=\; -322.71\; \backslash text\{\; J\}\; +\; \backslash text\{heat\; capacity\}\; \backslash times\; 15.2\; \backslash text\{\; °C\}$.

Rearranging this equation allows us to solve for the heat capacity of the calorimeter:

$heat\; capacity\}\; \backslash times\; 15.2\; \backslash text\{\; °C\}\; =\; 1,559,677.29\; \backslash text\{\; J\}$.

Finally, dividing both sides by the temperature change gives:

$heat\; capacity\}\; =\; \backslash frac\{1,559,677.29\; \backslash text\{\; J\}\}\{15.2\; \backslash text\{\; °C\}\}\; \backslash approx\; 1.03\; \backslash times\; 10^5\; \backslash text\{\; J/°C\}$.

This value represents the heat capacity of the bomb calorimeter, indicating its ability to absorb heat during the combustion process.