This is the amount of heat required (in kJ, 3 sig figs) to fully convert 3.12 kg of water from a liquid at 28
°C to steam at 112 °C.

To calculate the total amount of heat required to convert 3.12 kg of water from liquid at 28°C to steam at 112°C, we need to account for three stages of heat transfer:

1. **Heating the water from 28°C to 100°C (boiling point)** using the specific heat of liquid water.
2. **Converting liquid water at 100°C to steam** (phase change), which involves the heat of vaporization.
3. **Heating the steam from 100°C to 112°C** using the specific heat of steam.

We will use the following constants:
- Specific heat of water (\(C_{\text{water}}\)) = 4.18 J/g°C
- Heat of vaporization (\(L_{\text{vaporization}}\)) = 2260 J/g
- Specific heat of steam (\(C_{\text{steam}}\)) = 2.00 J/g°C

### Step 1: Heating water from 28°C to 100°C
The formula for heat transfer is:
\[
q_1 = m \cdot C_{\text{water}} \cdot \Delta T
\]
Where:
- \(m = 3120 \, \text{g}\) (since \(3.12 \, \text{kg} = 3120 \, \text{g}\))
- \(C_{\text{water}} = 4.18 \, \text{J/g°C}\)
- \(\Delta T = 100 - 28 = 72°C\)

\[
q_1 = 3120 \times 4.18 \times 72 = 939,456 \, \text{J} = 939.5 \, \text{kJ}
\]

### Step 2: Converting water at 100°C to steam
The formula for the heat of vaporization is:
\[
q_2 = m \cdot L_{\text{vaporization}}
\]
Where:
- \(L_{\text{vaporization}} = 2260 \, \text{J/g}\)

\[
q_2 = 3120 \times 2260 = 7,051,200 \, \text{J} = 7051.2 \, \text{kJ}
\]

### Step 3: Heating steam from 100°C to 112°C
The formula for heating the steam is:
\[
q_3 = m \cdot C_{\text{steam}} \cdot \Delta T
\]
Where:
- \(C_{\text{steam}} = 2.00 \, \text{J/g°C}\)
- \(\Delta T = 112 - 100 = 12°C\)

\[
q_3 = 3120 \times 2.00 \times 12 = 74,880 \, \text{J} = 74.9 \, \text{kJ}
\]

### Total Heat
Now, we sum all three amounts of heat:
\[
q_{\text{total}} = q_1 + q_2 + q_3 = 939.5 + 7051.2 + 74.9 = 8065.6 \, \text{kJ}
\]

The total amount of heat required is **8060 kJ** (rounded to 3 significant figures).

Would you like any further clarification or details? Here are 5 related questions for you:

1. How would the required heat change if the mass of water increased?
2. How do the specific heats of water and steam affect the energy calculation?
3. What is the difference between latent heat and sensible heat in this context?
4. What happens to the molecular structure of water during the phase change?
5. Can this method be applied to other substances, like alcohol or ammonia?

**Tip**: Always check the phase change temperature and the specific heat capacity for each state when performing heat transfer calculations.

This is the amount of heat required (in kJ, 3 sig figs) to fully convert 3.12 kg of water from a liquid at 28 °C to steam at 112 °C.

Learn how to calculate the total amount of heat required to convert 3.12 kg of water from 28°C to steam at 112°C. This solution covers thermodynamic concepts such as heat transfer, phase changes, and specific heat calculations. Suitable for high school students in Grades 10-12.

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