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To solve this problem, we need to calculate the total energy that must be released to convert 65.0 g of steam at 115.0°C to liquid water at 100.0°C. This involves two steps:

1. **Cooling the steam from 115.0°C to 100.0°C** (while it is still in the gaseous state).
2. **Condensing the steam at 100.0°C** into liquid water (phase change).

### Step 1: Cooling the steam

The formula to calculate the heat released when cooling steam is:
\[
q_1 = m \cdot C_{\text{steam}} \cdot \Delta T
\]
Where:
- \( q_1 \) = heat released during cooling (in joules, J)
- \( m \) = mass of steam (65.0 g)
- \( C_{\text{steam}} \) = specific heat capacity of steam, \( 2.02 \, \text{J/g}^\circ\text{C} \)
- \( \Delta T \) = change in temperature (final temperature - initial temperature)

For this step:
\[
\Delta T = 100.0^\circ \text{C} - 115.0^\circ \text{C} = -15.0^\circ \text{C}
\]
Now we can calculate \( q_1 \).

### Step 2: Condensing the steam

The formula to calculate the heat released during condensation is:
\[
q_2 = m \cdot \Delta H_{\text{vaporization}}
\]
Where:
- \( q_2 \) = heat released during condensation (in joules, J)
- \( \Delta H_{\text{vaporization}} \) = latent heat of vaporization of water, \( 2260 \, \text{J/g} \)
- \( m \) = mass of steam (65.0 g)

Now let's compute both steps.The total energy that must be released to cool 65.0 g of steam from 115.0°C to liquid water at 100.0°C is **144,930.5 J** (rounded to 3 significant digits: **1.45 × 10⁵ J**).

Would you like further clarification or details on any of the steps?

Here are 5 related questions to expand on this problem:
1. How does the specific heat capacity differ for water in its liquid and gas forms?
2. What is the process of condensation and why does it release energy?
3. How would the result change if the mass of the steam was different?
4. What would happen if the steam was cooled below 100.0°C?
5. Can we calculate the energy required to raise the temperature of liquid water from 100.0°C?

**Tip**: Remember that phase changes, like condensation, involve large amounts of energy compared to simply changing the temperature of a substance.

How much energy must be released from 65.0 g of steam at 115.0°C in order to turn it into liquid water at 100.0°C?

Learn how to calculate the energy released from 65.0 g of steam at 115.0°C to liquid water at 100.0°C using thermodynamics and phase change principles. This step-by-step solution covers heat transfer and phase change formulas. Suitable for high school students studying thermodynamics.

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