\(300 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\) is added to \(100 \mathrm{~g}\) of ice at…
\(300 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\) is added to \(100 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\). The final temperature of mixture will be
\(25^{\circ} \mathrm{C}\)
\(0^{\circ} \mathrm{C}\)
\(12.5^{\circ} \mathrm{C}\)
\(30^{\circ} \mathrm{C}\)
Solution
Given, mass of water
\(m_w=300 \mathrm{~g}\)
Temperature of water, \(T_w=25^{\circ} \mathrm{C}\)
Mass of ice,
\(\begin{gathered}
m_i=100 \mathrm{~g} \\
T_i=0^{\circ} \mathrm{C}
\end{gathered}\)
Heat required to convert the ice at \(0^{\circ} \mathrm{C}\) to water at \(0^{\circ} \mathrm{C}\)
\(\begin{array}{lll}
Q_1 & =m_i L_i=100 \times 80 \quad\left[L_i=80 \mathrm{cal} / \mathrm{g}\right] \\
& =8000 \mathrm{cal} \quad \ldots(\mathrm{i})
\end{array}\)
Heat given by water from \(25^{\circ} \mathrm{C}\) water to \(0^{\circ} \mathrm{C}\) water
\(\begin{aligned}
Q_2 & =m_w c \Delta T=300 \times 1 \times(25-0) \\
\Rightarrow & =7500 \mathrm{cal} \quad \ldots (ii)
\end{aligned}\)
From Eqs. (i) and (ii), we observed that
\(Q_2 < Q_1\)
Hence, total ice will not be melt, so final temperature of mixture will be \(0^{\circ} \mathrm{C}\) with some unmelted ice in mixture.