Calculate the amount of energy in joules that must be added to 245.0 g of ice at -45.0 ºC to turn it all into liquid water at 0.00 ºC.  Hint, two separate calculations must be done and then added together. how to solve  this

To calculate the amount of energy required to convert 245.0 g of ice at -45.0 ºC into liquid water at 0.00 ºC, we need to perform two separate calculations:

1. **Energy required to heat the ice from -45.0 ºC to 0.0 ºC** (This involves only heating the ice).
2. **Energy required to melt the ice at 0.0 ºC** (This involves the phase change from solid to liquid).

### Step 1: Heating the ice from -45.0 ºC to 0.0 ºC

To calculate the energy required to raise the temperature of ice, we use the formula:

\[
q_1 = m \cdot c_{\text{ice}} \cdot \Delta T
\]

Where:
- \( q_1 \) is the energy in joules (J).
- \( m \) is the mass of ice, which is 245.0 g.
- \( c_{\text{ice}} \) is the specific heat capacity of ice, approximately 2.09 J/g·ºC.
- \( \Delta T \) is the temperature change, which is:

\[
\Delta T = 0.0 - (-45.0) = 45.0 ºC
\]

Thus, the energy required to heat the ice:

\[
q_1 = 245.0 \, \text{g} \cdot 2.09 \, \text{J/g·ºC} \cdot 45.0 \, ºC
\]

### Step 2: Melting the ice at 0.0 ºC

For melting ice, we use the heat of fusion formula:

\[
q_2 = m \cdot L_f
\]

Where:
- \( q_2 \) is the energy in joules (J).
- \( m \) is the mass of ice, 245.0 g.
- \( L_f \) is the latent heat of fusion for ice, which is 334 J/g.

Thus, the energy required to melt the ice:

\[
q_2 = 245.0 \, \text{g} \cdot 334 \, \text{J/g}
\]

### Total Energy

Finally, to find the total energy required, sum both energies:

\[
q_{\text{total}} = q_1 + q_2
\]

Now, let’s calculate the values.### Calculation Results:
1. **Energy required to heat the ice from -45.0 ºC to 0.0 ºC:**  
   \( q_1 = 23,042.25 \, \text{J} \)
   
2. **Energy required to melt the ice at 0.0 ºC:**  
   \( q_2 = 81,830.0 \, \text{J} \)

### Total energy required:  
\[
q_{\text{total}} = 23,042.25 \, \text{J} + 81,830.0 \, \text{J} = 104,872.25 \, \text{J}
\]

Thus, **104,872.25 J** of energy is required to turn 245.0 g of ice at -45.0 ºC into liquid water at 0.00 ºC.

Would you like more details or have any questions?

Here are 5 related questions:
1. How much energy is needed to convert water at 100 ºC to steam?
2. How would the calculation change if the mass of ice were doubled?
3. What is the specific heat of water, and how does it compare to ice?
4. How does latent heat affect phase changes?
5. How does temperature change affect energy requirements in other materials?

**Tip:** Always ensure to differentiate between heating (temperature change) and phase change (like melting or boiling) when calculating energy.

Calculate the amount of energy in joules that must be added to 245.0 g of ice at -45.0 ºC to turn it all into liquid water at 0.00 ºC. Hint, two separate calculations must be done and then added together. How to solve this?

Learn how to calculate the total energy required to heat 245.0 g of ice from -45.0 ºC to liquid water at 0.00 ºC. This detailed step-by-step solution covers key thermodynamic concepts, specific heat capacity, and latent heat of fusion, suitable for high school students.

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