A 1.250 g sample of octane `(C_(8)H_(18))` is burned in excess of oxygen in a bomb calorimeter. The temperatre of the calorimeter rises from 294.05 K to 300.78K. If heat capacity of the calorimeter is 8.93kJ`//`K, find the heat transferred to the calorimeter. Also calculate the enthalpy combustion of the sample of octane.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the temperature change (ΔT)
We need to find the change in temperature of the calorimeter.

\[ \Delta T = T_{final} - T_{initial} \]

Given:
- \( T_{final} = 300.78 \, K \)
- \( T_{initial} = 294.05 \, K \)

Calculating ΔT:

\[ \Delta T = 300.78 \, K - 294.05 \, K = 6.73 \, K \]

### Step 2: Calculate the heat transferred to the calorimeter (q)
We use the formula:

\[ q = C \cdot \Delta T \]

Where:
- \( C \) = heat capacity of the calorimeter = 8.93 kJ/K
- \( \Delta T = 6.73 \, K \)

Substituting the values:

\[ q = 8.93 \, \text{kJ/K} \times 6.73 \, K = 60.00 \, \text{kJ} \]

### Step 3: Calculate the enthalpy of combustion per mole of octane
First, we need to find the number of moles of octane burned. The molar mass of octane \( (C_8H_{18}) \) is calculated as follows:

\[ \text{Molar mass of } C_8H_{18} = (8 \times 12.01) + (18 \times 1.008) = 114.22 \, g/mol \]

Now, we find the number of moles in the 1.250 g sample:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.250 \, g}{114.22 \, g/mol} = 0.01095 \, mol \]

Now, we can calculate the enthalpy of combustion per mole of octane:

\[ \Delta H_{combustion} = \frac{q}{\text{number of moles}} = \frac{60.00 \, kJ}{0.01095 \, mol} = 5485.4 \, kJ/mol \]

### Final Results
- Heat transferred to the calorimeter: **60.00 kJ**
- Enthalpy of combustion of octane: **5485.4 kJ/mol**

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