How much energy must be supplied to change 36 g of ice at `0^(@)C` to water at room temperature `25^(@)C`? `{:(,"Data for water",),(DeltaH_("fusion")^(@),,6.01kJ//mol),(C_(P" liquid"),,4.18JK^(-1)g^(-1)):}`

Options

A. 12 kJ
B. 16 kJ
C. 19 kJ
D. 22 kJ

Answer

To solve the problem of how much energy must be supplied to change 36 g of ice at 0°C to water at room temperature (25°C), we will follow these steps:

### Step 1: Calculate the number of moles of ice
The molar mass of ice (H₂O) is 18 g/mol.

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{36 \, \text{g}}{18 \, \text{g/mol}} = 2 \, \text{mol} \]

### Step 2: Calculate the energy required for melting the ice
The heat of fusion (ΔH_fusion) for water is given as 6.01 kJ/mol. To convert 2 moles of ice to water at 0°C:

\[ \text{Energy required} = \text{Number of moles} \times \Delta H_{fusion} = 2 \, \text{mol} \times 6.01 \, \text{kJ/mol} = 12.02 \, \text{kJ} \]

### Step 3: Calculate the energy required to heat the water from 0°C to 25°C
We will use the formula:

\[ q = m \cdot C \cdot \Delta T \]

Where:
- \( m = 36 \, \text{g} \) (mass of water)
- \( C = 4.18 \, \text{J/g·K} \) (specific heat capacity of water)
- \( \Delta T = T_{final} - T_{initial} = 25°C - 0°C = 25 \, \text{K} \)

Now, substituting the values:

\[ q = 36 \, \text{g} \cdot 4.18 \, \text{J/g·K} \cdot 25 \, \text{K} = 3765 \, \text{J} \]

Convert Joules to kilojoules:

\[ q = \frac{3765 \, \text{J}}{1000} = 3.765 \, \text{kJ} \]

### Step 4: Calculate the total energy required
Now, we add the energy required for melting the ice and the energy required to heat the water:

\[ \text{Total energy} = 12.02 \, \text{kJ} + 3.765 \, \text{kJ} = 15.785 \, \text{kJ} \]

### Step 5: Round off the total energy
Rounding off gives us approximately:

\[ \text{Total energy} \approx 16 \, \text{kJ} \]

### Final Answer
The total energy required to change 36 g of ice at 0°C to water at 25°C is approximately **16 kJ**.

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