How many grams of ice at \(-20^\circ\mathrm{C}\) will be needed to cool \(1.1\) litres of water from \(30^\circ\mathrm{C}\) to \(20^\circ\mathrm{C}\)?
\[
c_{\text{ice}}=0.5\ \text{cal g}^{-1}\,{}^{\circ}\mathrm{C}^{-1}
\]
\[
L_f=80\ \text{cal g}^{-1}
\]
Show Hint
When ice is added to water, include:
\[
\text{Heating of ice}
+
\text{Melting}
+
\text{Heating of melted water}.
\]
All three contributions are essential.
Step 1: Energy bookkeeping. The warm water gives up heat; the ice soaks it up in three jobs: warming the ice to $0^\circ$C, melting it, then warming the melt-water to $20^\circ$C. Set heat lost equal to heat gained.
Step 2: Heat released by the water. Mass of water $=1.1\times1000=1100$ g. Cooling from $30^\circ$C to $20^\circ$C, \[ Q_{\text{lost}}=1100\times1\times10=11000\ \text{cal}. \]
Step 3: Heat each gram of ice needs. Warm ice $-20^\circ$C to $0^\circ$C: $0.5\times20=10$ cal/g. Melt it: $80$ cal/g. Warm the water $0^\circ$C to $20^\circ$C: $1\times20=20$ cal/g.
Step 4: Add them up. Total per gram \[ =10+80+20=110\ \text{cal/g}. \]
Step 5: Solve for the mass of ice. If $m$ grams of ice are used, \[ m\times110=11000. \] So \[ m=\frac{11000}{110}=100\ \text{g}. \]
Step 6: State the answer. Exactly $100$ grams of ice are needed. \[ \boxed{100\ \text{g}} \]
