Water's heat of fusion is 80. cal/g, its specific heat is 1.0 g.°C, and its heat of vaporization is 540 cal/g •
A canister is filled with 360g of ice and 100. g of liquid water, both at 0 °C . The canister is placed in an oven until all the H2O has boiled off and the canister is empty.
How much energy in calories was absorbed?
Express your answer to two significant figures and include the appropriate units.

To calculate the total energy absorbed in this process, we need to account for three different phases:

1. **Melting the ice** (heat of fusion),
2. **Heating the water** from 0°C to 100°C (specific heat capacity),
3. **Vaporizing the water** (heat of vaporization).

### Step 1: Melting the ice
The energy required to melt ice is given by:
\[
Q_1 = m_{\text{ice}} \times \text{heat of fusion}
\]
Where:
- \( m_{\text{ice}} = 360 \, \text{g} \)
- Heat of fusion = \( 80 \, \text{cal/g} \)

\[
Q_1 = 360 \, \text{g} \times 80 \, \text{cal/g} = 28,800 \, \text{cal}
\]

### Step 2: Heating the water from 0°C to 100°C
The energy required to heat water is given by:
\[
Q_2 = (m_{\text{ice}} + m_{\text{water}}) \times \text{specific heat of water} \times \Delta T
\]
Where:
- \( m_{\text{ice}} + m_{\text{water}} = 360 \, \text{g} + 100 \, \text{g} = 460 \, \text{g} \)
- Specific heat of water = \( 1.0 \, \text{cal/g}^\circ \text{C} \)
- \( \Delta T = 100^\circ \text{C} - 0^\circ \text{C} = 100^\circ \text{C} \)

\[
Q_2 = 460 \, \text{g} \times 1.0 \, \text{cal/g}^\circ \text{C} \times 100^\circ \text{C} = 46,000 \, \text{cal}
\]

### Step 3: Vaporizing the water
The energy required to vaporize water is given by:
\[
Q_3 = (m_{\text{ice}} + m_{\text{water}}) \times \text{heat of vaporization}
\]
Where:
- Heat of vaporization = \( 540 \, \text{cal/g} \)

\[
Q_3 = 460 \, \text{g} \times 540 \, \text{cal/g} = 248,400 \, \text{cal}
\]

### Total Energy Absorbed
Now, summing up the energy required for all three steps:
\[
Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 28,800 \, \text{cal} + 46,000 \, \text{cal} + 248,400 \, \text{cal} = 323,200 \, \text{cal}
\]

To express this to two significant figures:
\[
Q_{\text{total}} \approx 3.2 \times 10^5 \, \text{cal}
\]

### Final Answer:
The total energy absorbed is **3.2 × 10⁵ calories**.

Would you like further explanation or details?

Here are 5 related questions for further exploration:
1. How would the calculation change if the initial temperature of the ice was -10°C?
2. What is the heat of fusion and why is it important for phase changes?
3. How much energy would be needed if the water were heated to 120°C in a pressurized system?
4. Can this process be reversed, and how would the energy calculations differ?
5. What role does specific heat play in determining the energy absorbed during temperature changes?

**Tip:** Remember that phase changes (like melting or boiling) involve energy without changing temperature.

Water's heat of fusion is 80. cal/g, its specific heat is 1.0 g.°C, and its heat of vaporization is 540 cal/g. A canister is filled with 360g of ice and 100g of liquid water, both at 0°C. The canister is placed in an oven until all the H2O has boiled off and the canister is empty. How much energy in calories was absorbed? Express your answer to two significant figures and include the appropriate units.

Calculate the total energy required to melt 360g of ice, heat 460g of water from 0°C to 100°C, and vaporize it. This thermodynamics problem explores phase changes and calorimetry concepts with heat of fusion and vaporization calculations. Suitable for students in Grades 11-12.

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