In a container of negligible mass 30g of steam at `100^@C` is added to 200g of water that has a temperature of `40^@C` If no heat is lost to the surroundings, what is the final temperature of the system? Also find masses of water and steam in equilibrium. Take `L_v = 539 cal//g and c_(water) = 1 cal//g-^@C.`

Explanation

To solve the problem, we will apply the principle of calorimetry, which states that in a closed system, the heat lost by one substance is equal to the heat gained by another substance.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of steam, \( m_s = 30 \, \text{g} \)
- Initial temperature of steam, \( T_s = 100^\circ C \)
- Mass of water, \( m_w = 200 \, \text{g} \)
- Initial temperature of water, \( T_w = 40^\circ C \)
- Latent heat of vaporization, \( L_v = 539 \, \text{cal/g} \)
- Specific heat of water, \( c_w = 1 \, \text{cal/g}^\circ C \)

2. **Calculate the Heat Required to Raise the Temperature of Water to 100°C:**
\[ Q_w = m_w \cdot c_w \cdot (T_f - T_w) \]
Here, \( T_f \) is the final temperature we want to find. The heat required to raise the temperature of water from \( 40^\circ C \) to \( 100^\circ C \) is:
\[ Q_w = 200 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot (100 - 40) = 200 \cdot 60 = 12000 \, \text{cal} \]

3. **Calculate the Heat Released by Steam When It Condenses:**
When steam condenses, it releases heat equal to the mass of steam converted to water multiplied by the latent heat of vaporization. Let \( m \) be the mass of steam that condenses:
\[ Q_s = m \cdot L_v \]
The heat released by the steam must equal the heat gained by the water:
\[ Q_s = Q_w \]
Thus, we have:
\[ m \cdot 539 = 12000 \]
Solving for \( m \):
\[ m = \frac{12000}{539} \approx 22.26 \, \text{g} \]

4. **Determine the Remaining Mass of Steam:**
The mass of steam remaining after condensation is:
\[ m_{remaining} = m_s - m = 30 \, \text{g} - 22.26 \, \text{g} \approx 7.74 \, \text{g} \]

5. **Calculate the Total Mass of Water at Final Temperature:**
The total mass of water in the system after the steam has condensed is:
\[ m_{total} = m_w + m = 200 \, \text{g} + 22.26 \, \text{g} = 222.26 \, \text{g} \]

6. **Final Temperature of the System:**
Since all the steam has turned into water at \( 100^\circ C \) and the remaining steam is at \( 100^\circ C \), the final temperature of the system will be \( 100^\circ C \).

### Final Answers:
- Final temperature of the system: \( 100^\circ C \)
- Mass of steam remaining: \( 7.74 \, \text{g} \)
- Mass of water (including condensed steam): \( 222.26 \, \text{g} \)