In the following equation calculate the value of `H`.
`1kg` syeam at `200^(@)C = H +1 kg` water at `100^(@)C (S_"system" =" Constant" = 0.5 cal//gm^(@)C)`

Explanation

To solve the problem, we need to calculate the total heat released (H) when 1 kg of steam at 200°C is converted to 1 kg of water at 100°C. This involves two processes: cooling the steam from 200°C to 100°C and then condensing the steam into water at 100°C.

### Step-by-step Solution:

1. **Identify the mass of steam**:
The mass of steam is given as 1 kg, which is equivalent to 1000 grams.

\[ m = 1000 \, \text{g} \]

2. **Cooling the steam from 200°C to 100°C**:
We will use the formula for heat transfer:
\[ Q_1 = m \cdot S \cdot \Delta T \]
where:
- \( S \) is the specific heat capacity of steam, given as \( 0.5 \, \text{cal/g°C} \).
- \( \Delta T \) is the change in temperature, which is \( 200°C - 100°C = 100°C \).

Plugging in the values:
\[ Q_1 = 1000 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 100°C \]
\[ Q_1 = 1000 \cdot 0.5 \cdot 100 = 50000 \, \text{cal} \]

3. **Condensing the steam at 100°C to water**:
The heat released during the phase change (condensation) is given by:
\[ Q_2 = m \cdot L \]
where:
- \( L \) is the latent heat of vaporization, given as \( 540 \, \text{cal/g} \).

Plugging in the values:
\[ Q_2 = 1000 \, \text{g} \cdot 540 \, \text{cal/g} \]
\[ Q_2 = 540000 \, \text{cal} \]

4. **Total heat released (H)**:
The total heat released (H) is the sum of the heat released during both processes:
\[ H = Q_1 + Q_2 \]
\[ H = 50000 \, \text{cal} + 540000 \, \text{cal} \]
\[ H = 590000 \, \text{cal} \]

5. **Convert to kilocalories**:
Since \( 1 \, \text{kcal} = 1000 \, \text{cal} \):
\[ H = \frac{590000 \, \text{cal}}{1000} = 590 \, \text{kcal} \]

### Final Answer:
The value of \( H \) is \( 590 \, \text{kcal} \).