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Question
85 g of water at 30°C is cooled to 5°C by adding certain mass of ice. Find the mass of ice required.
[Specific heat capacity of water = 4 .2 J g−1 °C−1, Specific latent heat of fusion = 336 J g−1]
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Solution
Given: Mass of water, \(m_1=85\,\text{g}\)
Initial temperature of water, \(T_1=30^\circ\text{C}\)
Final temperature, \(T_2=5^\circ\text{C}\)
Specific heat capacity of water, \(c=4.2\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Specific latent heat of fusion of ice, \(L=336\,\text{J g}^{-1}\)
Assuming the ice is initially at 0°C and no heat is lost to the surroundings,
According to the principle of calorimetry,
\[ \text{Heat lost by water}=\text{Heat gained by ice} \]
The heat lost by water.
\[ Q_1=m_1c(T_1-T_2) \]
\[ =85\times4.2\times(30-5) \]
\[ =85\times4.2\times25 \]
\[ =8925\,\text{J} \]
The heat gained by ice.
Ice at 0°C first melts into water and then the melted water warms to 5°C.
\[ Q_2=m_2L+m_2c(T_2-0) \]
\[ =336m_2+4.2m_2\times5 \]
\[ =336m_2+21m_2 \]
\[ =357m_2 \]
The mass of ice required.
According to the principle of calorimetry,
\[ Q_1=Q_2 \]
\[ 8925=357m_2 \]
\[ m_2=\frac{8925}{357} \]
\[ {m_2=25\,\text{g}} \]
