The heat capactiy 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 `0^(@)` C, each containing 9.0 g of the of water, needed to cool 500 g of liquid water from `20^(@)`C to `0^(@)`C ?

Options

A. 1
B. 7
C. 14
D. None of these

Answer

To solve the problem, we need to determine how many ice cubes are required to cool 500 g of liquid water from 20°C to 0°C. Each ice cube contains 9.0 g of water at 0°C.

### Step-by-Step Solution:

1. **Convert the mass of water to moles**:
The molar mass of water (H₂O) is approximately 18 g/mol.
\[ \text{Number of moles of water} = \frac{500 \, \text{g}}{18 \, \text{g/mol}} \approx 27.78 \, \text{mol} \]

2. **Calculate the heat lost by the liquid water**:
The heat lost by the water can be calculated using the formula:
\[ q = n \cdot C \cdot \Delta T \]
where:
- \( n \) = number of moles of water = 27.78 mol
- \( C \) = molar heat capacity of water = 75.6 J/mol·K
- \( \Delta T \) = change in temperature = 20°C - 0°C = 20 K

Substituting the values:
\[ q = 27.78 \, \text{mol} \cdot 75.6 \, \text{J/mol·K} \cdot 20 \, \text{K} \]
\[ q \approx 27.78 \cdot 75.6 \cdot 20 \approx 41964.48 \, \text{J} \]

3. **Convert the heat lost to kJ**:
\[ q \approx 41.96 \, \text{kJ} \]

4. **Calculate the heat gained by the ice cubes**:
Each ice cube contains 9.0 g of water, which is:
\[ \text{Number of moles in one ice cube} = \frac{9.0 \, \text{g}}{18 \, \text{g/mol}} = 0.5 \, \text{mol} \]
The enthalpy of fusion of ice is given as 6.0 kJ/mol. Therefore, the heat gained by one ice cube when it melts is:
\[ q_{\text{ice}} = n \cdot \Delta H_{\text{fusion}} = 0.5 \, \text{mol} \cdot 6.0 \, \text{kJ/mol} = 3.0 \, \text{kJ} \]

5. **Determine the number of ice cubes needed**:
Let \( N \) be the number of ice cubes. The total heat gained by \( N \) ice cubes is:
\[ q_{\text{total}} = N \cdot 3.0 \, \text{kJ} \]
Setting the heat lost by the water equal to the heat gained by the ice cubes:
\[ 41.96 \, \text{kJ} = N \cdot 3.0 \, \text{kJ} \]
Solving for \( N \):
\[ N = \frac{41.96 \, \text{kJ}}{3.0 \, \text{kJ}} \approx 13.99 \]

6. **Round up to the nearest whole number**:
Since we cannot have a fraction of an ice cube, we round up to the nearest whole number:
\[ N = 14 \]

### Final Answer:
The smallest number of ice cubes needed is **14**.