To find the amount of heat energy required to freeze 4.5 g of water at 6°C to ice at 0°C, we can break down the problem into two parts:
1. **Cooling the water from 6°C to 0°C**: We will use the formula for heat transfer due to a temperature change, which is given by:
\[
Q_1 = m \cdot s \cdot \Delta T
\]
where:
- \( Q_1 \) is the heat lost during cooling,
- \( m \) is the mass of the water (in kg),
- \( s \) is the specific heat capacity of water,
- \( \Delta T \) is the change in temperature.
2. **Freezing the water at 0°C**: We will use the formula for heat transfer during a phase change, which is given by:
\[
Q_2 = m \cdot L
\]
where:
- \( Q_2 \) is the heat lost during freezing,
- \( L \) is the latent heat of fusion of water.
### Step 1: Calculate the heat lost while cooling the water
Given:
- Mass of water, \( m = 4.5 \, \text{g} = 4.5 / 1000 \, \text{kg} = 0.0045 \, \text{kg} \)
- Specific heat capacity of water, \( s = 41900 \, \text{J/kg°C} \)
- Initial temperature, \( T_i = 6°C \)
- Final temperature, \( T_f = 0°C \)
The change in temperature, \( \Delta T = T_f - T_i = 0 - 6 = -6°C \).
Now, substituting the values into the formula:
\[
Q_1 = 0.0045 \, \text{kg} \cdot 41900 \, \text{J/kg°C} \cdot (-6°C)
\]
Calculating \( Q_1 \):
\[
Q_1 = 0.0045 \cdot 41900 \cdot (-6) = -1131.3 \, \text{J}
\]
### Step 2: Calculate the heat lost while freezing the water
Given:
- Latent heat of fusion, \( L = 3.33 \times 10^5 \, \text{J/kg} \)
Now, substituting the values into the formula:
\[
Q_2 = 0.0045 \, \text{kg} \cdot 3.33 \times 10^5 \, \text{J/kg}
\]
Calculating \( Q_2 \):
\[
Q_2 = 0.0045 \cdot 3.33 \times 10^5 = 1498.5 \, \text{J}
\]
### Step 3: Total heat lost
The total heat energy required to freeze the water is the sum of the heat lost during cooling and the heat lost during freezing:
\[
Q_{\text{total}} = Q_1 + Q_2
\]
Substituting the values:
\[
Q_{\text{total}} = (-1131.3 \, \text{J}) + (-1498.5 \, \text{J}) = -2629.8 \, \text{J}
\]
Since we are interested in the magnitude of heat energy required, we take the positive value:
\[
Q_{\text{total}} = 2629.8 \, \text{J}
\]
### Final Answer:
The amount of heat energy required to freeze 4.5 g of water at 6°C to ice at 0°C is **2629.8 J**.
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