To solve for the value of \( H \) in the equation given, we need to analyze the heat transfer involved in converting 1 kg of steam at \( 200^\circ C \) to 1 kg of water at \( 100^\circ C \).
### Step-by-Step Solution:
1. **Identify the Components of the Problem**:
- We have 1 kg of steam at \( 200^\circ C \).
- We need to find the heat \( H \) that is required to convert 1 kg of water at \( 100^\circ C \) to steam at \( 200^\circ C \).
2. **Convert Steam at \( 200^\circ C \) to Steam at \( 100^\circ C \)**:
- First, we need to cool the steam from \( 200^\circ C \) to \( 100^\circ C \).
- The heat lost during this process can be calculated using the formula:
\[
Q_1 = m \cdot c \cdot \Delta T
\]
- Where:
- \( m = 1000 \, \text{g} \) (mass of steam)
- \( c = 0.5 \, \text{cal/g}^\circ C \) (specific heat capacity of steam)
- \( \Delta T = 200^\circ C - 100^\circ C = 100^\circ C \)
- Plugging in the values:
\[
Q_1 = 1000 \, \text{g} \cdot 0.5 \, \text{cal/g}^\circ C \cdot 100^\circ C = 50000 \, \text{cal}
\]
3. **Convert Steam at \( 100^\circ C \) to Water**:
- Next, we need to condense the steam at \( 100^\circ C \) to water at \( 100^\circ C \).
- The latent heat of vaporization for water is approximately \( 540 \, \text{cal/g} \).
- The heat released during this phase change is:
\[
Q_2 = m \cdot L
\]
- Where:
- \( L = 540 \, \text{cal/g} \)
- Plugging in the values:
\[
Q_2 = 1000 \, \text{g} \cdot 540 \, \text{cal/g} = 540000 \, \text{cal}
\]
4. **Calculate Total Heat \( H \)**:
- The total heat \( H \) required to convert 1 kg of water at \( 100^\circ C \) to steam at \( 200^\circ C \) is the sum of the heat lost by the steam as it cools and condenses:
\[
H = Q_1 + Q_2
\]
- Therefore:
\[
H = 50000 \, \text{cal} + 540000 \, \text{cal} = 590000 \, \text{cal}
\]
5. **Convert to kcal**:
- Since \( 1 \, \text{kcal} = 1000 \, \text{cal} \):
\[
H = \frac{590000 \, \text{cal}}{1000} = 590 \, \text{kcal}
\]
### Final Answer:
\[
H = 590 \, \text{kcal}
\]
