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Question
104g of water at 30°C is taken in a calorimeter made of copper of mass 42 g. When a certain mass of ice at 0°C is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10°C. Find the mass of ice added. [Specific heat capacity of water = 4.2 J g–1 °C–1 ; Specific latent heat of fusion of ice = 336 J g–1; Specific heat capacity of copper = 0.4 J g–1 °C–1] .
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Solution
Given: Mass of water, \(m_1=104\,\text{g}\)
Mass of copper calorimeter, \(m_2=42\,\text{g}\)
Initial temperature of water and calorimeter, \(T_1=30^\circ\text{C}\)
Initial temperature of ice = \(0^\circ\text{C}\)
Final temperature, \(T_2=10^\circ\text{C}\)
Specific heat capacity of water, \(c_1=4.2\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Specific heat capacity of copper, \(c_2=0.4\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Specific latent heat of fusion of ice, \(L=336\,\text{J g}^{-1}\)
According to the principle of calorimetry,
\[ \text{Heat lost}=\text{Heat gained} \]
The heat lost by water.
\[ Q_1=m_1c_1(T_1-T_2) \]
\[ =104\times4.2\times(30-10) \]
\[ =8736\,\text{J} \]
The heat lost by the copper calorimeter.
\[ Q_2=m_2c_2(T_1-T_2) \]
\[ =42\times0.4\times(30-10) \]
\[ =336\,\text{J} \]
The total heat lost.
\[ Q=Q_1+Q_2 \]
\[ =8736+336 \]
\[ =9072\,\text{J} \]
The heat gained by ice.
Ice at 0°C first melts into water and then the melted water warms to 10°C.
\[ Q=mL+mc_1(T_2-0) \]
\[ =336m+4.2m\times10 \]
\[ =336m+42m \]
\[ =378m \]
According to the principle of calorimetry,
\[ 9072=378m \]
\[ m=\frac{9072}{378} \]
\[{m=24\,\text{g}} \]
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