A copper calorimeter of mass 190 g contains 300 g of water at `0^(@)C` and 50 g of ice at `0^(@)C`.Find the amount of steam at `100^(@)C` required to raise the temperature of this mixture by `10^(@)C`. Given that the specific heat of copperis `420J//kg^(@)C`, latent heat of vaporization of water is `2.25xx10^(5)xxJ//kg` and latent heat of fusion of ice is `3.36xx10^(5)J//kg`.

Answer

To solve the problem step by step, we will calculate the heat absorbed by the ice and water mixture, the calorimeter, and then determine how much steam is required to provide that heat.

### Step 1: Calculate the heat absorbed by the ice to melt into water.
The heat absorbed by the ice (Q1) can be calculated using the formula:
\[ Q_1 = m \cdot L_f \]
where:
- \( m = 50 \, \text{g} = 0.050 \, \text{kg} \) (mass of ice)
- \( L_f = 3.36 \times 10^5 \, \text{J/kg} \) (latent heat of fusion)

Calculating Q1:
\[ Q_1 = 0.050 \, \text{kg} \times 3.36 \times 10^5 \, \text{J/kg} = 16800 \, \text{J} \]

### Step 2: Calculate the heat absorbed by the water to raise its temperature from 0°C to 10°C.
The total mass of water after melting the ice is:
\[ m_{water} = 300 \, \text{g} + 50 \, \text{g} = 350 \, \text{g} = 0.350 \, \text{kg} \]

The heat absorbed by the water (Q2) can be calculated using the formula:
\[ Q_2 = m \cdot c_w \cdot \Delta T \]
where:
- \( c_w = 4182 \, \text{J/kg°C} \) (specific heat capacity of water)
- \( \Delta T = 10 - 0 = 10 \, \text{°C} \)

Calculating Q2:
\[ Q_2 = 0.350 \, \text{kg} \times 4182 \, \text{J/kg°C} \times 10 \, \text{°C} = 14637 \, \text{J} \]

### Step 3: Calculate the heat absorbed by the copper calorimeter.
The heat absorbed by the calorimeter (Q3) can be calculated using the formula:
\[ Q_3 = m_{cal} \cdot c_{Cu} \cdot \Delta T \]
where:
- \( m_{cal} = 190 \, \text{g} = 0.190 \, \text{kg} \) (mass of the calorimeter)
- \( c_{Cu} = 420 \, \text{J/kg°C} \) (specific heat capacity of copper)

Calculating Q3:
\[ Q_3 = 0.190 \, \text{kg} \times 420 \, \text{J/kg°C} \times 10 \, \text{°C} = 798 \, \text{J} \]

### Step 4: Calculate the total heat absorbed by the mixture.
The total heat absorbed (Q_total) is the sum of Q1, Q2, and Q3:
\[ Q_{total} = Q_1 + Q_2 + Q_3 \]
Calculating Q_total:
\[ Q_{total} = 16800 \, \text{J} + 14637 \, \text{J} + 798 \, \text{J} = 32635 \, \text{J} \]

### Step 5: Calculate the mass of steam required to provide this heat.
The heat released by the steam when it condenses can be given by:
\[ Q_{steam} = m_{steam} \cdot L_v \]
where:
- \( L_v = 2.25 \times 10^6 \, \text{J/kg} \) (latent heat of vaporization)

Setting \( Q_{steam} = Q_{total} \):
\[ m_{steam} \cdot 2.25 \times 10^6 \, \text{J/kg} = 32635 \, \text{J} \]

Solving for \( m_{steam} \):
\[ m_{steam} = \frac{32635 \, \text{J}}{2.25 \times 10^6 \, \text{J/kg}} \approx 0.0145 \, \text{kg} \]

### Step 6: Convert the mass of steam to grams.
\[ m_{steam} = 0.0145 \, \text{kg} \times 1000 \, \text{g/kg} = 14.5 \, \text{g} \]

Thus, the amount of steam required is approximately **14.5 g**.