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Answer :
Total energy: 6170.57 J. Steps: Heat ice, melt ice, heat water, vaporize water, heat steam.
To calculate the energy required to convert a substance through multiple phases, we need to consider the energy required for each phase transition and for heating or cooling within each phase.
Here are the steps to solve this problem:
1. Calculate the energy required to raise the temperature of the ice from -12.0°C to its melting point (0°C).
2. Calculate the energy required to melt the ice at 0°C.
3. Calculate the energy required to raise the temperature of the resulting water from 0°C to its boiling point (100°C).
4. Calculate the energy required to vaporize the water at 100°C.
5. Calculate the energy required to raise the temperature of the resulting steam from 100°C to 105°C.
Let's start with the calculations:
1. Energy to raise the temperature of ice:
[tex]\[ q_1 = m \cdot c_{\text{ice}} \cdot \Delta T \][/tex]
[tex]\[ q_1 = 1.70 \, \text{g} \times 2.09 \, \text{J/g\°C} \times (0 - (-12)) \, \text{\°C} \][/tex]
[tex]\[ q_1 = 1.70 \, \text{g} \times 2.09 \, \text{J/g\°C} \times 12 \, \text{\°C} \][/tex]
[tex]\[ q_1 = 1.70 \, \text{g} \times 2.09 \, \text{J/g\°C} \times 12 \, \text{\°C} \][/tex]
[tex]\[ q_1 = 42.588 \, \text{J} \][/tex]
2. Energy to melt the ice:
[tex]\[ q_2 = m \cdot \Delta H_{\text{fusion}} \][/tex]
[tex]\[ q_2 = 1.70 \, \text{g} \times 334 \, \text{J/g} \][/tex]
[tex]\[ q_2 = 567.8 \, \text{J} \][/tex]
3. Energy to raise the temperature of water:
[tex]\[ q_3 = m \cdot c_{\text{water}} \cdot \Delta T \][/tex]
[tex]\[ q_3 = 1.70 \, \text{g} \times 4.18 \, \text{J/g\°C} \times (100 - 0) \, \text{\°C} \][/tex]
[tex]\[ q_3 = 1.70 \, \text{g} \times 4.18 \, \text{J/g\°C} \times 100 \, \text{\°C} \][/tex]
\[ q_3 = 710.92 \, \text{J} \]
4. Energy to vaporize the water:
[tex]\[ q_4 = m \cdot \Delta H_{\text{vap}} \][/tex]
[tex]\[ q_4 = 1.70 \, \text{g} \times 2260 \, \text{J/g} \][/tex]
[tex]\[ q_4 = 3832 \, \text{J} \][/tex]
5. Energy to raise the temperature of steam:
[tex]\[ q_5 = m \cdot c_{\text{steam}} \cdot \Delta T \][/tex]
[tex]\[ q_5 = 1.70 \, \text{g} \times 2.03 \, \text{J/g\°C} \times (105 - 100) \, \text{\°C} \][/tex]
[tex]\[ q_5 = 1.70 \, \text{g} \times 2.03 \, \text{J/g\°C} \times 5 \, \text{\°C} \][/tex]
[tex]\[ q_5 = 17.255 \, \text{J} \][/tex]
Now, sum up all these energies to find the total energy required:
[tex]\[ \text{Total energy} = q_1 + q_2 + q_3 + q_4 + q_5 \][/tex]
[tex]\[ \text{Total energy} = 42.588 \, \text{J} + 567.8 \, \text{J} + 710.92 \, \text{J} + 3832 \, \text{J} + 17.255 \, \text{J} \][/tex]
[tex]\[ \text{Total energy} = 6170.565 \, \text{J} \][/tex]
So, the total energy required to convert 1.70 g of ice originally at –12.0°C into steam at 105°C is approximately [tex]\( 6170.57 \, \text{J} \).[/tex]
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