Example

Question:

Calculate the heat required to convert 3 kg of ice at \(-12^\circ\mathrm{C}\) kept in a calorimeter to steam at \(100^\circ\mathrm{C}\) at atmospheric pressure.
Given:
Specific heat capacity of ice = \(2100\,\mathrm{J\,kg^{-1}\,K^{-1}}\)
Specific heat capacity of water = \(4186\,\mathrm{J\,kg^{-1}\,K^{-1}}\)
Latent heat of fusion of ice = \(3.35 \times 10^5\,\mathrm{J\,kg^{-1}}\)
Latent heat of steam = \(2.256 \times 10^6\,\mathrm{J\,kg^{-1}}\)

Solution:

Mass of ice, \(m = 3\,\mathrm{kg}\)

1. Heat required to convert ice at \(-12^\circ\mathrm{C}\) to ice at \(0^\circ\mathrm{C}\): \[ Q_1 = m s_\mathrm{ice} \Delta T_1 = 3 \times 2100 \times [0 - (-12)] = 75600\,\mathrm{J} \]
2. Heat required to melt ice at \(0^\circ\mathrm{C}\) to water: \[ Q_2 = m L_\mathrm{ice} = 3 \times 3.35 \times 10^5 = 1005000\,\mathrm{J} \]
3. Heat required to convert water at \(0^\circ\mathrm{C}\) to water at \(100^\circ\mathrm{C}\): \[ Q_3 = m s_\mathrm{water} \Delta T_2 = 3 \times 4186 \times (100 - 0) = 1255800\,\mathrm{J} \]
4. Heat required to convert water at \(100^\circ\mathrm{C}\) to steam at \(100^\circ\mathrm{C}\): \[ Q_4 = m L_\mathrm{steam} = 3 \times 2.256 \times 10^6 = 6768000\,\mathrm{J} \]
Total heat required: \[ Q = Q_1 + Q_2 + Q_3 + Q_4 = 75600 + 1005000 + 1255800 + 6768000 = 9.1 \times 10^6\,\mathrm{J} \]