# Calorimetry¶

## Units and Symbols¶

Symbol | Units | |
---|---|---|

Amount of heat transferred | \(q\) | \(J\) |

Mass of the substance | \(m\) | \(g\) |

Temperature change | \(\Delta T\) | \(^\circ C\) or \(^\circ K\) |

Heat capacity - Increase in a fixed body by 1°C | \(C\) | \(J\cdot^\circ C^{-1}\) |

Specific heat capacity - Increase in 1g of the substance by 1°C | \(s\) | \(J\cdot g^{-1} \cdot ^\circ C^{-1}\) |

## Equation - Heat Change¶

This equation and page is about heat transfer through **temperature** change.

Where:

- \(\Delta T=T_f−T_i\)

\[
q=ms\Delta T \hskip{1cm}\text{or}\hskip{1cm}q=C\Delta T
\]

## Calorimeters¶

### Bomb Calorimeter - Isolated System¶

- Closed bomb is immersed in a very specific amount of water
- Heat capacity of the calorimeter is known
- Sample is ignited
- Change of temperature in the water is recorded

### Constant Pressure Calorimeter - Open System¶

- Used for non combustion reactions
- Measures the change in the temperature of a solution

### Calculating a Value for q¶

\[
q_{system}=q_{reaction}+q_{water}+q_{bomb}
\]

- In the bomb calorimeter, as it’s an isolated system, the q_system=0, so energy is conserved throughtout \(q_{reaction}\), \(q_{water}\), and \(q_{bomb}\)
- Values for \(s\) and \(C\) are provided in documentation

Energy used to go from solid to liquid = latent heat of fusion Energy used to go from liquid to gas = latent heat of vaporisation

### Negatives¶

- A negative value indicates an exothermic reaction, as energy from the system is released to the surrounds
- A positive value indicates an endothermic reaction, as energy from the surrounds is taken by the system

E.g.

Calculate the amount of heat absorbed by 425 grams of water that is heated from −17°C to 12°C. The latent heat of fusion for water is 6.01 \(kJ\cdot mol^{-1}\).

**Step 1 and 3 - Heating from −17°C to 0°C and from 0°C to 12°C**

\[
\begin{align}
ms\Delta T&=q\\
425×4.184×(12−(−17))&=52.1\:kJ
\end{align}
\]

**Step 2 - melting at 0°C**

\[
\begin{align}
\frac{m}{M}&=n\\
\frac{425}{18.016}&=23.59\:mol\\
23.59\:mol×6.01kJ\cdot mol^{-1}&=141.8\:kJ\\
142+52.1&=193\:kJ
\end{align}
\]