In a mixture of `35 g` of ice and `35 g` of water in equilibrium, `4 gm` steam is passed. The whole mixture is in a copper calorimeter of mass `50 g`. Find the equilibrium temperature of the mixture. Given that specific heat of water is `4200` J/kg K and that of copper is `420` J/Kg K and latent heat of fusion of ice is `3.36xx10^(5)` J/kg and latent heat of vaporization of water is `2.25xx10^(6)` J/kg.

Answer

To find the equilibrium temperature of the mixture after passing 4 g of steam into a calorimeter containing 35 g of ice and 35 g of water, we will follow these steps:

### Step 1: Identify Initial Conditions
- Mass of ice, \( m_{ice} = 35 \, g = 0.035 \, kg \)
- Mass of water, \( m_{water} = 35 \, g = 0.035 \, kg \)
- Mass of steam, \( m_{steam} = 4 \, g = 0.004 \, kg \)
- Mass of copper calorimeter, \( m_{copper} = 50 \, g = 0.050 \, kg \)

### Step 2: Calculate Energy Released by Steam
The steam at 100 °C will first condense into water at 100 °C, releasing energy equal to its latent heat of vaporization.

\[ Q_{steam} = m_{steam} \times L_{v} \]
Where \( L_{v} = 2.25 \times 10^6 \, J/kg \)

\[ Q_{steam} = 0.004 \, kg \times 2.25 \times 10^6 \, J/kg = 9000 \, J = 9 \, kJ \]

### Step 3: Calculate Energy Required to Melt Ice
Next, we calculate the energy required to melt the ice at 0 °C into water at 0 °C.

\[ Q_{melt} = m_{ice} \times L_{f} \]
Where \( L_{f} = 3.36 \times 10^5 \, J/kg \)

\[ Q_{melt} = 0.035 \, kg \times 3.36 \times 10^5 \, J/kg = 11760 \, J = 11.76 \, kJ \]

### Step 4: Calculate Energy Released by Cooling Steam Water to 0 °C
The steam, after condensing, will cool down to 0 °C, releasing additional energy.

\[ Q_{cool} = m_{steam} \times c_{water} \times \Delta T \]
Where \( c_{water} = 4200 \, J/kg \cdot K \) and \( \Delta T = 100 \, °C - 0 \, °C = 100 \, K \)

\[ Q_{cool} = 0.004 \, kg \times 4200 \, J/kg \cdot K \times 100 \, K = 1680 \, J = 1.68 \, kJ \]

### Step 5: Total Energy Released by Steam
Total energy released by the steam:

\[ Q_{total} = Q_{steam} + Q_{cool} = 9 \, kJ + 1.68 \, kJ = 10.68 \, kJ \]

### Step 6: Compare Energy Released and Energy Required
The energy required to melt all the ice is 11.76 kJ, but the total energy released by the steam is only 10.68 kJ. This means not all ice will melt.

### Step 7: Calculate the Mass of Ice that Melts
Let \( m \) be the mass of ice that melts. The energy used to melt \( m \) grams of ice is given by:

\[ Q_{used} = m \times L_{f} \]

Setting \( Q_{used} = 10.68 \, kJ \):

\[ 10.68 \times 10^3 = m \times 3.36 \times 10^5 \]

Solving for \( m \):

\[ m = \frac{10.68 \times 10^3}{3.36 \times 10^5} = 0.0318 \, kg = 31.8 \, g \]

### Step 8: Determine Final Masses
- Remaining ice = \( 35 \, g - 31.8 \, g = 3.2 \, g \)
- Total water after melting = \( 35 \, g + 31.8 \, g + 4 \, g = 70.8 \, g \)

### Step 9: Final Composition
The final mixture consists of:
- Ice: \( 3.2 \, g \)
- Water: \( 70.8 \, g \)

### Step 10: Final Temperature of the Mixture
Since the remaining ice and water are both at 0 °C, the equilibrium temperature of the mixture will be:

\[ T_{final} = 0 \, °C \]

### Summary
The equilibrium temperature of the mixture after passing 4 g of steam into the calorimeter is \( 0 \, °C \).

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