Calorimetry means measurement of heat. When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the colder body, provided no heat is allowed to escape to the surroundings.

Calorimeter: A device in which heat measurement can be done is called a calorimeter.

A simple calorimeter is a vessel that is generally made up of copper with a stirrer of the same material. The vessel is kept in a wooden box to separate it thermally from the surrounding. A thermometer is used for the measurement of the temperature of the contents of the calorimeter. There is one opening through which a thermometer can be inserted to measure the change in thermal properties inside.

Objects which are at different temperatures are made to come in touch with each other in the calorimeter. In a calorimeter a fixed amount of fuel is burned, leading to the heating of water. Heat which is lost by the fuel is equal to the heat which is gained by the water. As a result, heat is exchanged between the objects and the calorimeter. Neglecting any heat exchange with the surrounding. This is why it is important that the calorimeter should be insulated from the environment; to improve the accuracy of the experiment. This exchange of heat is measured through the thermometer.

Calorimeter Principle

When two bodies of different temperatures (ideally a solid and a liquid) come into physical contact with one other, heat is transferred from the body with higher temperature to the body with lower temperature until thermal equilibrium is reached between them. The body at higher temperatures emits heat, whereas the body at lower temperatures absorbs heat. 

The principle of calorimetry explains the law of conservation of energy, which states that the total heat lost by the hot body equals the total heat gained by the cool body, that is:

Heat Lost = Heat Gained

The heat is calculated using the Calorimetry formula, 

\(
Q=m S {\Delta T}
\)
where \(Q\) is the heat capacity, \(m\) is the mass, \(S\) is the specific heat capacity and \(\Delta T\) is the temperature change.

Example 1: \(\mathrm{~A}\) sphere of \(0.047 \mathrm{~kg}\) aluminium is placed for sufficient time in a vessel containing bolling water, so that the sphere is at \(100^{\circ} \mathrm{C}\). It is then immediately transfered to \(0.14 \mathrm{~kg}\) copper calorimeter containing \(0.25 \mathrm{~kg}\) water at \(20{ }^{\circ} \mathrm{C}\). The temperature of water rises and attains a steady state at \(23^{\circ} \mathrm{C}\). Calculate the specific heat capacity of aluminium.

Solution:

In solving this example, we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter.
Mass of aluminium sphere \(\left(m_1\right)=0.047 \mathrm{~kg}\) Initial temperature of aluminium sphere \(=100{ }^{\circ} \mathrm{C}\) Final temperature \(=23^{\circ} \mathrm{C}\) Change in temperature \((\Delta T)=\left(100^{\circ} \mathrm{C}-23^{\circ} \mathrm{C}\right)=77^{\circ} \mathrm{C}\) Let specific heat capacity of aluminium be \(s_{\mathrm{Al}}\).
The amount of heat lost by the aluminium sphere \(=m_1 s_{A l} \Delta T=0.047 \mathrm{~kg} \times s_{A l} \times 77^{\circ} \mathrm{C}\)
Mass of water \(\left(m_2\right)=0.25 \mathrm{~kg}\)
Mass of calorimeter \(\left(m_3\right)=0.14 \mathrm{~kg}\)
Inttal temperature of water and calorimeter \(=20^{\circ} \mathrm{C}\)
Final temperature of the mixture \(=23^{\circ} \mathrm{C}\)
Change in temperature \(\left(\Delta T_2\right)=23^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=3{ }^{\circ} \mathrm{C}\)
Specific heat capacity of water \(\left(s_w\right)\)
\(
=4.18 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}
\)
Specific heat capacity of copper calorimeter \(=0.386 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\)
The amount of heat gained by water and calorimeter \(=m_2 s_{\mathrm{w}} \Delta T_2+m_3 s_{\mathrm{cu}} \Delta T_2\)
\(=\left(m_2 s_{\mathrm{w}}+m_3 s_{\mathrm{cu}}\right)\left(\Delta T_2\right)\)
\(=\left(0.25 \mathrm{~kg} \times 4.18 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}+0.14 \mathrm{~kg} \times\right.\)
\(\left.0.386 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)\left(23^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}\right)\)
In the steady state heat lost by the aluminium
sphere \(=\) heat gained by water \(+\) heat gained by calorimeter.
So, \(0.047 \mathrm{~kg} \times \mathrm{s}_{\mathrm{Al}} \times 77^{\circ} \mathrm{C}\)
\(=\left(0.25 \mathrm{~kg} \times 4.18 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}+0.14 \mathrm{~kg} \times\right.\)
\(\left.0.386 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)\left(3{ }^{\circ} \mathrm{C}\right)\)
\(s_{A l}=0.911 \mathrm{~kJ} \mathrm{~kg}{ }^{-1} \mathrm{~K}^{-1}\)