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)`
What is the final temperature of the system ?

Options

A. `48^(@)`C
B. `72^(@)`C
C. `94^(@)`C
D. `100^(@)`C

Explanation

To solve the problem, we need to consider the heat exchanges occurring in the system involving ice, steam, and water. The final temperature of the system will be determined by the heat gained by the ice and water and the heat lost by the steam.

Let's denote:
- Mass of ice, \( m_{ice} = 200 \, \text{g} \)
- Mass of steam, \( m_{steam} = 100 \, \text{g} \)
- Mass of water, \( m_{water} = 200 \, \text{g} \)
- Initial temperature of ice, \( T_{ice} = 0^\circ C \)
- Initial temperature of steam, \( T_{steam} = 100^\circ C \)
- Initial temperature of water, \( T_{water} = 55^\circ C \)

### Step 1: Heat gained by the ice
The ice will first melt and then the resulting water will heat up. The heat gained by the ice can be calculated as follows:

1. **Melting the ice**:
\[ Q_{melt} = m_{ice} \cdot L_f = 200 \, \text{g} \cdot 80 \, \text{cal/g} = 16000 \, \text{cal} \]

2. **Heating the melted ice (water) from \(0^\circ C\) to \(T\)**:
\[ Q_{heat\_ice} = m_{ice} \cdot s_w \cdot (T - 0) = 200 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot T = 200T \, \text{cal} \]

Thus, the total heat gained by the ice is:
\[ Q_{ice} = Q_{melt} + Q_{heat\_ice} = 16000 + 200T \, \text{cal} \]

### Step 2: Heat gained by the water
The water will also gain heat as it warms up from \(55^\circ C\) to \(T\):
\[ Q_{water} = m_{water} \cdot s_w \cdot (T - 55) = 200 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot (T - 55) = 200(T - 55) \, \text{cal} \]

### Step 3: Heat lost by the steam
The steam will condense and then the resulting water will cool down to \(T\):
1. **Condensing the steam**:
\[ Q_{condense} = m_{steam} \cdot L_v = 100 \, \text{g} \cdot 540 \, \text{cal/g} = 54000 \, \text{cal} \]

2. **Cooling the resulting water from \(100^\circ C\) to \(T\)**:
\[ Q_{cool\_steam} = m_{steam} \cdot s_w \cdot (100 - T) = 100 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot (100 - T) = 100(100 - T) \, \text{cal} \]

Thus, the total heat lost by the steam is:
\[ Q_{steam} = Q_{condense} + Q_{cool\_steam} = 54000 + 100(100 - T) \, \text{cal} \]

### Step 4: Setting up the heat balance equation
Since no heat is lost to the surroundings, the heat gained by the ice and water must equal the heat lost by the steam:
\[ Q_{ice} + Q_{water} = Q_{steam} \]
Substituting the equations we derived:
\[ (16000 + 200T) + 200(T - 55) = 54000 + 100(100 - T) \]

### Step 5: Simplifying the equation
Expanding and simplifying:
\[ 16000 + 200T + 200T - 11000 = 54000 + 10000 - 100T \]
\[ 16000 + 400T - 11000 = 64000 - 100T \]
\[ 5000 + 400T = 64000 - 100T \]
\[ 500T = 59000 \]
\[ T = \frac{59000}{500} = 118^\circ C \]

### Step 6: Final temperature
Since the final temperature cannot exceed \(100^\circ C\) (the boiling point of water), we must conclude that the final temperature of the system will stabilize at \(100^\circ C\).

### Final Answer
The final temperature of the system is \(100^\circ C\).