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प्रश्न
A piece of ice of mass 60 g is dropped into 140 g of water at 50°C. Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding.)
Specific heat capacity of water = 4.2 \(\text{J g}^{-1} k^{-1}\)
Specific latent heat of fusion of ice = 336\(\text{ J g}^{-1}\)
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उत्तर
Given: Mass of ice, \(m_1=60\,\text{g}\)
Mass of water, \(m_2=140\,\text{g}\)
Initial temperature of water = \(50^\circ\text{C}\)
Specific heat capacity of water, \(c=4.2\,\text{J g}^{-1}\,\text{K}^{-1}\)
Specific latent heat of fusion of ice, \(L=336\,\text{J g}^{-1}\)
Assumption: The ice is initially at 0°C.
According to the principle of calorimetry,
\[\text{Heat lost by hot water}=\text{Heat gained by ice} \]
The heat lost by hot water.
\[ Q_1=m_2c(50-T) \]
\[ Q_1=140\times4.2\times(50-T) \]
\[ Q_1=588(50-T) \]
The heat gained by ice.
Ice at 0°C first melts into water and then the melted water warms to the final temperature \(T\).
\[ Q_2=m_1L+m_1cT \]
\[ Q_2=60\times336+60\times4.2\times T \]
\[ Q_2=20160+252T \]
Apply the principle of calorimetry.
\[ Q_1=Q_2 \]
\[ 588(50-T)=20160+252T \]
\[ 29400-588T=20160+252T \]
\[ 29400-20160=588T+252T \]
\[ 9240=840T \]
\[ T=\frac{9240}{840} \]
\[ {T=11^\circ\text{C}} \]
