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The heat capacity of liquid water is $75.6 J / mol K,$ , while the enthalpy of fusion of ice is $6.0 kJ / mol$ . What is the smallest number of ice cubes at $O^{\circ} C$ , each containing $9.0 g$ of water, needed to cool 500 g of liquid water from $20^{\circ} C$ to $O^{\circ} C$ ?

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Detailed Solution

Heat released by water $= \frac{500}{18} \times 75.6 \times 20$

Let n number of cubes are required.

So, $n \times \frac{9}{18} \times 6000 = \frac{500}{18} \times 75.6 \times 20$

$n = 14$ .

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