Question

The temperature of 3.50 kg of water is raised by 1.17 when 1g of hydrazine N2H4 is burned in a bomb calorimeter. The calorimeter has a heat capacity of 883 how much heat is given off by the sample

Answer 1

The heat given off by the sample is approximately 16842.19 J.

To calculate the heat given off, we can use the formula
`\( Q = mc\Delta T \)`, where Q is the heat, m is the mass, c is the specific heat capacity, and
`\( \Delta T \)` is the change in temperature.

Given:

- Mass of water (m) = 3.50 kg

- Change in temperature (
`\( \Delta T \)`) = 1.17 °C

- Specific heat capacity of water (c) ≈ 4186 J/kg°C

- Heat capacity of the calorimeter (
`\( C_{\text{calorimeter}} \)`) = 883 J/°C

`\[ Q_{\text{water}} = m \cdot c \cdot \Delta T = 3.50 \, \text{kg} \cdot 4186 \, \text{J/kg C} \cdot 1.17 \,{^\circ C} \]`

`\[ Q_{\text{calorimeter}} = C_{\text{calorimeter}} \cdot \Delta T = 883 \, \text{J/ C} \cdot 1.17 \°C} \]`

The total heat given off (
`\( Q_{\text{total}} \)`) is the sum of
`\( Q_{\text{water}} \) and \( Q_{\text{calorimeter}} \).`

`\[ Q_{\text{total}} = Q_{\text{water}} + Q_{\text{calorimeter}} \]`

`\[ Q_{\text{total}} \approx (3.50 \, \text{kg} \cdot 4186 \, \text{J/kgC} \cdot 1.17 \, \text{C}) + (883 \, \text{J/C} \cdot 1.17 \, \text{C}) \]`

`\[ Q_{\text{total}} \approx 16842.19 \, \text{J} \]`