The amount of heat energy required to freeze 4.5 g of water of `6^(@)C` to ice at `0^(@)C` is `[S=41900 J kg^(-1)K^(-1), L=3.33xx10^(5)Jkg^(-1)]`

Options

A. 1612 J
B. 1512 J
C. 1132 J
D. 1499 J

Answer

To solve the problem of calculating the amount of heat energy required to freeze 4.5 g of water at 6°C to ice at 0°C, we will follow these steps:

### Step 1: Convert mass from grams to kilograms
Given mass \( m = 4.5 \, \text{g} \).

To convert grams to kilograms:
\[ m = \frac{4.5 \, \text{g}}{1000} = 0.0045 \, \text{kg} \]

### Step 2: Calculate the heat required to cool the water from 6°C to 0°C
We will use the formula for heat transfer due to a temperature change:
\[ Q_s = m \cdot S \cdot \Delta T \]
Where:
- \( S = 41900 \, \text{J/kg/K} \) (specific heat of water)
- \( \Delta T = T_1 - T_2 = 6°C - 0°C = 6 \, \text{K} \)

Substituting the values:
\[ Q_s = 0.0045 \, \text{kg} \cdot 41900 \, \text{J/kg/K} \cdot 6 \, \text{K} \]
\[ Q_s = 0.0045 \cdot 41900 \cdot 6 = 1134.3 \, \text{J} \]

### Step 3: Calculate the heat required for the phase change from water to ice
For the phase change, we use the latent heat of fusion:
\[ Q_L = m \cdot L \]
Where:
- \( L = 3.33 \times 10^5 \, \text{J/kg} \) (latent heat of fusion)

Substituting the values:
\[ Q_L = 0.0045 \, \text{kg} \cdot 3.33 \times 10^5 \, \text{J/kg} \]
\[ Q_L = 0.0045 \cdot 3.33 \times 10^5 = 1498.5 \, \text{J} \]

### Step 4: Calculate the total heat energy required
The total heat energy \( Q \) required is the sum of the heat lost while cooling and the heat lost during the phase change:
\[ Q = Q_s + Q_L \]
\[ Q = 1134.3 \, \text{J} + 1498.5 \, \text{J} = 2632.8 \, \text{J} \]

### Final Answer
The total amount of heat energy required to freeze 4.5 g of water at 6°C to ice at 0°C is approximately:
\[ \boxed{2632.8 \, \text{J}} \]