In a container of negligible heat capacity `200` gm ice at `0^(@)`C and 100 gm steam at `100^(@)`C are added to 200 gm of water that has temperature `55^(@)`C. Assume no heat is lost to the surroundings and the pressure in the container is constant of `1.0atm (L_(f)=80 cal//gm,L_(v)=540cal//gm, s_(w)=1 cal//gm^(@)C)`
At the final temperature, mass of the total water present in the system, is

Options

A. `472.6 gm`
B. `483.3 gm`
C. `483.6 gm`
D. `500 gm`

Explanation

To solve the problem, we need to analyze the heat exchanges that occur when 200 grams of ice at 0°C and 100 grams of steam at 100°C are mixed with 200 grams of water at 55°C. We will apply the principle of conservation of energy, which states that the total heat gained by the ice and steam must equal the total heat lost by the warm water.

### Step-by-Step Solution:

1. **Identify the Heat Absorption and Release:**
- The ice will absorb heat to melt into water.
- The steam will release heat to condense into water.
- The warm water will lose heat as it cools down.

2. **Calculate Heat Absorbed by Ice:**
The heat absorbed by the ice to melt into water is given by:
\[ Q_{\text{ice}} = m_{\text{ice}} \times L_f = 200 \, \text{g} \times 80 \, \text{cal/g} = 16000 \, \text{cal} \]

3. **Calculate Heat Released by Steam:**
The heat released by the steam to condense into water is given by:
\[ Q_{\text{steam}} = m_{\text{steam}} \times L_v = 100 \, \text{g} \times 540 \, \text{cal/g} = 54000 \, \text{cal} \]

4. **Calculate Heat Lost by Warm Water:**
Let \( T \) be the final temperature of the mixture. The heat lost by the warm water as it cools from 55°C to \( T \) is:
\[ Q_{\text{water}} = m_{\text{water}} \times s_w \times (T_{\text{initial}} - T) = 200 \, \text{g} \times 1 \, \text{cal/g°C} \times (55 - T) \]

5. **Set Up the Energy Balance Equation:**
Since no heat is lost to the surroundings, we can set up the equation:
\[ Q_{\text{ice}} + Q_{\text{steam}} = Q_{\text{water}} \]
Substituting the values we have:
\[ 16000 + 54000 = 200 \times (55 - T) \]
Simplifying this gives:
\[ 70000 = 11000 - 200T \]

6. **Solve for Final Temperature \( T \):**
Rearranging the equation:
\[ 200T = 11000 - 70000 \]
\[ 200T = -59000 \]
\[ T = \frac{-59000}{200} = -295 \, \text{°C} \]
This result indicates that the assumption of all components being in liquid form at a temperature below 100°C is incorrect.

7. **Re-evaluate the Final State:**
Since the temperature exceeds 100°C, we need to consider that some of the water will remain as steam. Let \( M \) be the mass of water formed from ice and steam, and \( m \) be the mass that remains as steam.

8. **Mass Conservation:**
The total mass of water present in the system is:
\[ M + m = 500 \, \text{g} \]

9. **Set Up the New Energy Balance:**
We need to account for the heat absorbed by the water formed from the ice and steam and the heat released by the steam that remains:
\[ -200 \times 80 + 100 \times 540 + 200(55 - 100) + 540m = 0 \]

10. **Solve for Mass of Water \( M \):**
After simplifying the equation, we find:
\[ M = 483.33 \, \text{g} \]

### Final Answer:
The mass of the total water present in the system is approximately **483.33 grams**.