Question

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 \(20^\circ\mathrm{C}\)? \[ c_{\text{ice}}=0.5\ \text{cal g}^{-1}\,{}^{\circ}\mathrm{C}^{-1} \] \[ L_f=80\ \text{cal g}^{-1} \]

• A. \(90\) g
• B. \(90.1\) g
• C. \(100.1\) g
• D. \(100\) g

Hint:

When ice is added to water, include: \[ \text{Heating of ice} + \text{Melting} + \text{Heating of melted water}. \] All three contributions are essential.

Answer 1

The Correct Option is D

Concept: Heat lost by water = Heat gained by ice.

Step 1: Calculate heat lost by water. Mass of water: \[ m_w=1.1\times1000=1100\ \text{g}. \] Heat lost: \[ Q_w = 1100\times1\times(30-20) \] \[ = 11000\ \text{cal}. \]

Step 2: Calculate heat required per gram of ice. Heating ice from \(-20^\circ\mathrm{C}\) to \(0^\circ\mathrm{C}\): \[ Q_1 = 0.5\times20 = 10\ \text{cal/g}. \] Melting ice: \[ Q_2 = 80\ \text{cal/g}. \] Heating melted water from \(0^\circ\mathrm{C}\) to \(20^\circ\mathrm{C}\): \[ Q_3 = 20\ \text{cal/g}. \] Total heat absorbed per gram: \[ Q=10+80+20 \] \[ =110\ \text{cal/g}. \]

Step 3: Find the mass of ice required. \[ m = \frac{11000}{110} \] \[ =100\ \text{g}. \] \[\begin{aligned} \boxed{100\ \text{g}} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.