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To calculate the energy released when converting 500 grams of water at 15°C to ice at -20°C, we need to consider the following steps of the process:
1. Cooling the water from 15°C to 0°C.
2. Freezing the water at 0°C to ice.
3. Cooling the ice from 0°C to -20°C.
Each step involves a specific calculation for energy, and we'll sum these energies to find the total energy released.
### Step 1: Cooling the Water from 15°C to 0°C
We first calculate the energy required to cool the water from its initial temperature (15°C) to the freezing point (0°C).
The formula we use is:
[tex]\[ q_1 = m \cdot c_{\text{water}} \cdot \Delta T \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the water (500 grams)
- [tex]\( c_{\text{water}} \)[/tex] is the specific heat capacity of water (4.186 J/g°C)
- [tex]\( \Delta T \)[/tex] is the change in temperature (0°C - 15°C)
Substituting the values:
[tex]\[ q_1 = 500 \, \text{g} \times 4.186 \, \text{J/g°C} \times (0 - 15) \, \text{°C} \][/tex]
[tex]\[ q_1 = 500 \times 4.186 \times -15 \][/tex]
[tex]\[ q_1 = -31395 \, \text{J} \][/tex]
### Step 2: Freezing the Water at 0°C to Ice
Next, we calculate the energy released when the water freezes at 0°C.
The formula we use is:
[tex]\[ q_2 = m \cdot L_f \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the water (500 grams)
- [tex]\( L_f \)[/tex] is the heat of fusion for water (334 J/g)
Substituting the values:
[tex]\[ q_2 = 500 \, \text{g} \times 334 \, \text{J/g} \][/tex]
[tex]\[ q_2 = 167000 \, \text{J} \][/tex]
### Step 3: Cooling the Ice from 0°C to -20°C
Finally, we calculate the energy required to cool the ice from 0°C to -20°C.
The formula we use is:
[tex]\[ q_3 = m \cdot c_{\text{ice}} \cdot \Delta T \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the ice (500 grams)
- [tex]\( c_{\text{ice}} \)[/tex] is the specific heat capacity of ice (2.09 J/g°C)
- [tex]\( \Delta T \)[/tex] is the change in temperature (-20°C - 0°C)
Substituting the values:
[tex]\[ q_3 = 500 \, \text{g} \times 2.09 \, \text{J/g°C} \times (-20) \, \text{°C} \][/tex]
[tex]\[ q_3 = 500 \times 2.09 \times -20 \][/tex]
[tex]\[ q_3 = -20900 \, \text{J} \][/tex]
### Total Energy Released
To find the total energy released, we sum the energy values from each step.
[tex]\[ \text{Total energy} = q_1 + q_2 + q_3 \][/tex]
[tex]\[ \text{Total energy} = -31395 \, \text{J} + 167000 \, \text{J} + -20900 \, \text{J} \][/tex]
[tex]\[ \text{Total energy} = 114705 \, \text{J} \][/tex]
Therefore, the total energy released in converting 500 grams of water at 15°C into ice at -20°C is 114705 joules.
1. Cooling the water from 15°C to 0°C.
2. Freezing the water at 0°C to ice.
3. Cooling the ice from 0°C to -20°C.
Each step involves a specific calculation for energy, and we'll sum these energies to find the total energy released.
### Step 1: Cooling the Water from 15°C to 0°C
We first calculate the energy required to cool the water from its initial temperature (15°C) to the freezing point (0°C).
The formula we use is:
[tex]\[ q_1 = m \cdot c_{\text{water}} \cdot \Delta T \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the water (500 grams)
- [tex]\( c_{\text{water}} \)[/tex] is the specific heat capacity of water (4.186 J/g°C)
- [tex]\( \Delta T \)[/tex] is the change in temperature (0°C - 15°C)
Substituting the values:
[tex]\[ q_1 = 500 \, \text{g} \times 4.186 \, \text{J/g°C} \times (0 - 15) \, \text{°C} \][/tex]
[tex]\[ q_1 = 500 \times 4.186 \times -15 \][/tex]
[tex]\[ q_1 = -31395 \, \text{J} \][/tex]
### Step 2: Freezing the Water at 0°C to Ice
Next, we calculate the energy released when the water freezes at 0°C.
The formula we use is:
[tex]\[ q_2 = m \cdot L_f \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the water (500 grams)
- [tex]\( L_f \)[/tex] is the heat of fusion for water (334 J/g)
Substituting the values:
[tex]\[ q_2 = 500 \, \text{g} \times 334 \, \text{J/g} \][/tex]
[tex]\[ q_2 = 167000 \, \text{J} \][/tex]
### Step 3: Cooling the Ice from 0°C to -20°C
Finally, we calculate the energy required to cool the ice from 0°C to -20°C.
The formula we use is:
[tex]\[ q_3 = m \cdot c_{\text{ice}} \cdot \Delta T \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the ice (500 grams)
- [tex]\( c_{\text{ice}} \)[/tex] is the specific heat capacity of ice (2.09 J/g°C)
- [tex]\( \Delta T \)[/tex] is the change in temperature (-20°C - 0°C)
Substituting the values:
[tex]\[ q_3 = 500 \, \text{g} \times 2.09 \, \text{J/g°C} \times (-20) \, \text{°C} \][/tex]
[tex]\[ q_3 = 500 \times 2.09 \times -20 \][/tex]
[tex]\[ q_3 = -20900 \, \text{J} \][/tex]
### Total Energy Released
To find the total energy released, we sum the energy values from each step.
[tex]\[ \text{Total energy} = q_1 + q_2 + q_3 \][/tex]
[tex]\[ \text{Total energy} = -31395 \, \text{J} + 167000 \, \text{J} + -20900 \, \text{J} \][/tex]
[tex]\[ \text{Total energy} = 114705 \, \text{J} \][/tex]
Therefore, the total energy released in converting 500 grams of water at 15°C into ice at -20°C is 114705 joules.
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