If 452,000 Joules of heat energy is required to completely evaporate a certain amount of water at 100°C, how many grams of water is being evaporated? (Latent heat of vaporization for water = 2260 J/g)

To find out how many grams of water are being evaporated, we can use the formula:

\[
Q = m \cdot L
\]

Where:
- \( Q \) is the heat energy required (in Joules),
- \( m \) is the mass of the water (in grams),
- \( L \) is the latent heat of vaporization (in Joules per gram).

Given:
- \( Q = 452,000 \, \text{J} \),
- \( L = 2260 \, \text{J/g} \).

Rearranging the formula to solve for \( m \):

\[
m = \frac{Q}{L}
\]

Substituting the values:

\[
m = \frac{452,000}{2260} = 200 \, \text{grams}
\]

Therefore, 200 grams of water is being evaporated.

Would you like any further details or explanations? 

Here are 5 related questions:
1. What is latent heat, and why is it important in phase changes?
2. How does the latent heat of vaporization compare to latent heat of fusion?
3. What happens to the temperature of water during evaporation at 100°C?
4. How is the specific heat capacity different from latent heat?
5. How can you calculate the energy required to raise the temperature of water before vaporization?

**Tip:** Latent heat is energy absorbed or released during a phase change without changing temperature.

Discover how to calculate the amount of water evaporated when 452,000 Joules of heat energy is applied at 100°C. This detailed explanation covers the formula Q = m * L, using the latent heat of vaporization for water (2260 J/g). Ideal for students in Grades 9-12.

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