Example

Question:

A sphere of 0.047 kg aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transferred to a 0.14 kg copper calorimeter containing 0.25 kg water at 20°C. The temperature of water rises and attains a steady state at 23°C. Calculate the specific heat capacity of aluminium.

Solution:

At steady state, heat given by aluminium = heat absorbed by water and calorimeter.
Mass of aluminium sphere (\(m_1\)) = 0.047 kg.
Initial temperature (\(T_1\)) = 100°C, Final temperature = 23°C.
Change in temperature: \(\Delta T = 100 - 23 = 77^\circ\mathrm{C}\).
Specific heat capacity of aluminium = \(s_\mathrm{Al}\).

Heat lost by aluminium: \[ Q_{\mathrm{Al}} = m_1 s_\mathrm{Al} \Delta T = 0.047 \times s_\mathrm{Al} \times 77 \]
Mass of water (\(m_2\)) = 0.25 kg.
Mass of calorimeter (\(m_3\)) = 0.14 kg.
Initial temperature = 20°C, Final = 23°C.
\(\Delta T_2 = 23 - 20 = 3^\circ\mathrm{C}\).

Specific heat capacity of water (\(s_\mathrm{w}\)) = \(4.18 \times 10^3\,\mathrm{J\,kg}^{-1}\,\mathrm{K}^{-1}\).
Specific heat capacity of copper calorimeter (\(s_\mathrm{cu}\)) = \(0.386 \times 10^3\,\mathrm{J\,kg}^{-1}\,\mathrm{K}^{-1}\).
Heat gained: \[ Q_{\mathrm{gain}} = (m_2 s_\mathrm{w} + m_3 s_\mathrm{cu}) \Delta T_2 \] \[ = [0.25 \times 4.18 \times 10^3 + 0.14 \times 0.386 \times 10^3] \times 3 \]
At steady state: \[ 0.047 \times s_\mathrm{Al} \times 77 = (0.25 \times 4.18 \times 10^3 + 0.14 \times 0.386 \times 10^3) \times 3 \] \[ s_\mathrm{Al} = 0.911\,\mathrm{kJ\,kg}^{-1}\,\mathrm{K}^{-1} \]