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प्रश्न
A copper vessel of mass 100 g contains 150 g of water at 50°C. How much ice is needed to cool it to 5°C?
Given: Specific heat capacity of copper = 0.4 Jg−1 °C−1
The Specific heat capacity of water = 4.2 Jg−1 °C−1
The Specific latent heat of fusion ice = 336 Jg−1
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उत्तर
Given: Mass of copper vessel, \(m_1=100\,\text{g}\)
Mass of water, \(m_2=150\,\text{g}\)
Initial temperature, \(T_1=50^\circ\text{C}\)
Final temperature, \(T_2=5^\circ\text{C}\)
Specific heat capacity of copper, \(c_1=0.4\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Specific heat capacity of water, \(c_2=4.2\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Specific latent heat of fusion of ice, \(L=336\,\text{J g}^{-1}\)
Assumption: Ice is initially at 0°C, and no heat is lost to the surroundings.
Now, heat energy lost by the copper vessel is:
\[ Q_1=m_1c_1(T_1-T_2) \]
\[ =100\times0.4\times(50-5) \]
\[ =1800\,\text{J} \]
Similarly, heat energy lost by water is:
\[ Q_2=m_2c_2(T_1-T_2) \]
\[ =150\times4.2\times(50-5) \]
\[ =28350\,\text{J} \]
Hence, the total heat energy lost is:
\[ Q=Q_1+Q_2 \]
\[ =1800+28350=30150\,\text{J} \]
Ice at 0°C melts and the resulting water warms to 5°C.
\[ Q=mL+mc_2(5-0) \]
\[ =336m+4.2m\times5 \]
\[ =357m \]
According to the principle of calorimetry,
\[ \text{Heat lost}=\text{Heat gained} \]
\[ 30150=357m \]
\[ m=\frac{30150}{357}\]
\[=84.45\,\text{g} \]
संबंधित प्रश्न
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If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy. The change in temperature continues till the temperatures of both objects attain the same value. In this process, the cold object gains heat energy and the hot object loses heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat-resistant box then no energy can flow from inside the box or come into the box. In this situation, we get the following principle.
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