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To solve this problem, we need to determine how much heat is absorbed when a sample of water is heated from 60.0°C to steam at 140.0°C. This process includes two stages:

1. Heating the water from 60.0°C to 100.0°C (its boiling point).
2. Converting the water at 100.0°C to steam (this involves the latent heat of vaporization).
3. Heating the steam from 100.0°C to 140.0°C.

Let's go step by step:

### Step 1: Heating the water from 60.0°C to 100.0°C

The formula to calculate the heat absorbed to raise the temperature of a substance is:

\[
q_1 = m \cdot c \cdot \Delta T
\]

Where:
- \( q_1 \) is the heat absorbed in this stage.
- \( m \) is the mass of the water.
- \( c \) is the specific heat capacity of water ( \( c = 4.18 \, \text{J/g°C} \) for water).
- \( \Delta T \) is the temperature change.

Given:
- Mass of water, \( m = 2.00 \times 10^2 \, \text{g} = 200 \, \text{g} \).
- Initial temperature, \( T_i = 60.0°C \).
- Final temperature for this stage, \( T_f = 100.0°C \).
- \( \Delta T = T_f - T_i = 100.0°C - 60.0°C = 40.0°C \).

Thus,

\[
q_1 = 200 \, \text{g} \times 4.18 \, \text{J/g°C} \times 40.0°C
\]

Let's calculate this.

\[
q_1 = 200 \times 4.18 \times 40 = 33440 \, \text{J}
\]

### Step 2: Converting water at 100.0°C to steam at 100.0°C

The heat required to convert water into steam at 100.0°C is given by:

\[
q_2 = m \cdot H_v
\]

Where:
- \( q_2 \) is the heat absorbed during the phase change.
- \( m \) is the mass of the water (which remains 200 g or 0.2 kg).
- \( H_v \) is the latent heat of vaporization (given as \( 3.34 \times 10^5 \, \text{J/kg} \)).

Thus,

\[
q_2 = 0.2 \, \text{kg} \times 3.34 \times 10^5 \, \text{J/kg} = 66800 \, \text{J}
\]

### Step 3: Heating the steam from 100.0°C to 140.0°C

The formula to calculate the heat absorbed to raise the temperature of steam is similar to Step 1:

\[
q_3 = m \cdot c_{\text{steam}} \cdot \Delta T
\]

Where:
- \( q_3 \) is the heat absorbed to raise the temperature of steam.
- \( c_{\text{steam}} = 2.01 \, \text{J/g°C} \) is the specific heat capacity of steam.
- \( \Delta T = 140.0°C - 100.0°C = 40.0°C \).

Thus,

\[
q_3 = 200 \, \text{g} \times 2.01 \, \text{J/g°C} \times 40.0°C
\]

Let's calculate this:

\[
q_3 = 200 \times 2.01 \times 40 = 16080 \, \text{J}
\]

### Step 4: Total Heat Absorbed

Now, we sum up the heat absorbed in all three steps:

\[
q_{\text{total}} = q_1 + q_2 + q_3
\]

Substitute the values:

\[
q_{\text{total}} = 33440 \, \text{J} + 66800 \, \text{J} + 16080 \, \text{J}
\]

Now, calculate the total:

\[
q_{\text{total}} = 116320 \, \text{J}
\]

### Final Answer:

The total heat absorbed is **116,320 J** or **116.32 kJ**.

---

Would you like any further details or have any questions? Here are some related questions to consider:

1. What is the specific heat capacity and how does it affect the amount of heat absorbed?
2. How does the latent heat of vaporization impact phase changes?
3. Why does heating steam require less energy than boiling water?
4. What other factors can influence the amount of heat absorbed by a substance?
5. How would the calculations change if the mass of water was different?

**Tip:** Always remember to convert units appropriately, especially when using values in joules and kilograms.

A 2.00 × 10² g sample of water at 60.0°C is heated to steam at 140.0°C. How much heat is absorbed? Hv = 3.34 × 10⁵ J/kg.

This problem explains how to calculate the heat absorbed when a 2.00 × 10² g sample of water is heated from 60.0°C to steam at 140.0°C, including latent heat of vaporization and specific heat capacities. Suitable for students in Grades 10-12.

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