Stearic acid `[CH_(3)(CH_(2))_(16)CO_(2)H]` is a fatty acid the part of fat that stores most of the energy .1.0 g of Stearic acid was burnt in a bomb calorimeter . The bomb had capacity of 652 `J//^(@)C`. If the temperature of 500 g water rose from 25.0 to `39.3^(@)C` how much heat is released when the stearic acid was burned?
`["Given "C_(p)(H_(2)O)=4.18J//g^(@)c]`

Options

A. 39.21 kJ
B. 29.91 kJ
C. 108 kJ
D. 9.32 kJ

Explanation

To solve the problem of how much heat is released when 1.0 g of stearic acid is burned in a bomb calorimeter, we can follow these steps:

### Step 1: Calculate the heat absorbed by the water.
The formula to calculate the heat absorbed by the water is given by:
\[ q_{\text{water}} = m \cdot C_p \cdot \Delta T \]
Where:
- \( m \) = mass of water = 500 g
- \( C_p \) = specific heat capacity of water = 4.18 J/g°C
- \( \Delta T \) = change in temperature = final temperature - initial temperature = \( 39.3°C - 25.0°C = 14.3°C \)

Substituting the values:
\[ q_{\text{water}} = 500 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot 14.3 \, \text{°C} \]
\[ q_{\text{water}} = 500 \cdot 4.18 \cdot 14.3 = 29885 \, \text{J} \]

### Step 2: Calculate the heat absorbed by the bomb calorimeter.
The formula to calculate the heat absorbed by the bomb calorimeter is:
\[ q_{\text{calorimeter}} = C_{\text{calorimeter}} \cdot \Delta T \]
Where:
- \( C_{\text{calorimeter}} \) = capacity of the bomb calorimeter = 652 J/°C
- \( \Delta T \) = same change in temperature = 14.3°C

Substituting the values:
\[ q_{\text{calorimeter}} = 652 \, \text{J/°C} \cdot 14.3 \, \text{°C} \]
\[ q_{\text{calorimeter}} = 652 \cdot 14.3 = 9335.6 \, \text{J} \]

### Step 3: Calculate the total heat released.
The total heat released when stearic acid is burned is the sum of the heat absorbed by the water and the calorimeter:
\[ q_{\text{total}} = q_{\text{water}} + q_{\text{calorimeter}} \]
Substituting the values:
\[ q_{\text{total}} = 29885 \, \text{J} + 9335.6 \, \text{J} = 39220.6 \, \text{J} \]

### Step 4: Convert the heat released to kilojoules.
To convert joules to kilojoules, we divide by 1000:
\[ q_{\text{total}} = \frac{39220.6 \, \text{J}}{1000} = 39.22 \, \text{kJ} \]

### Conclusion
The amount of heat released when 1.0 g of stearic acid is burned is approximately **39.22 kJ**.

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