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प्रश्न
A calorimeter of mass 50 g and specific heat capacity \[0.42\ \text{J g}^{-1}{}^{\circ}\text{C}^{-1}\] contains some mass of water at \[20^{\circ}\text{C}\]. A metal piece of mass 20 g at \[100^{\circ}\text{C}\] is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be \[22^{\circ}\text{C}\]. Find the mass of water used in the calorimeter.
[Specific heat capacity of metal = \[0.3\ \text{J g}^{-1}{}^{\circ}\text{C}^{-1}\] and specific heat capacity of water = \[4.2\ \text{J g}^{-1}{}^{\circ}\text{C}^{-1}\]]
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उत्तर
Given: Mass of calorimeter, \(m_1=50\,\text{g}\)
Specific heat capacity of calorimeter, \(c_1=0.42\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Mass of metal piece, \(m_2=20\,\text{g}\)
Specific heat capacity of metal, \(c_2=0.3\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Specific heat capacity of water, \(c_3=4.2\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Initial temperature of water and calorimeter = \(20^\circ\text{C}\)
Initial temperature of metal = \(100^\circ\text{C}\)
Final temperature = \(22^\circ\text{C}\)
According to the principle of calorimetry,
\[ \text{Heat lost by metal}=\text{Heat gained by water and calorimeter} \]
The heat lost by the metal piece.
\[ Q_1=m_2c_2\Delta T \]
\[ =20\times0.3\times(100-22) \]
\[ =20\times0.3\times78 \]
\[ =468\,\text{J} \]
The heat gained by the calorimeter.
\[ Q_2=m_1c_1\Delta T \]
\[ =50\times0.42\times(22-20) \]
\[ =42\,\text{J} \]
The heat gained by water.
\[ Q_3=mc_3\Delta T \]
\[ =m\times4.2\times(22-20) \]
\[ =m\times4.2\times2\]
\[ =8.4m\,\text{J} \]
The mass of water.
According to the principle of calorimetry,
\[ Q_1=Q_2+Q_3 \]
\[ 468=42+8.4m \]
\[ 8.4m=468-42 \]
\[ 8.4m=426 \]
\[ m=\frac{426}{8.4} \]
\[{m=50.71\,\text{g}} \]
