A 2 kg copper block is heated to `500^@C` and then it is placed on a large block of ice at `0^@C`. If the specific heat capacity of copper is `400 "J/kg/"^@C` and latent heat of fusion of water is `3.5xx 10^5` J/kg. The amount of ice that can melt is :

Options

A. (7/8) kg
B. (7/5) kg
C. (8/7) kg
D. (5/7) kg

Explanation

To solve the problem step by step, we will use the principle of conservation of energy. The heat lost by the copper block will be equal to the heat gained by the ice.

### Step 1: Calculate the heat lost by the copper block

The formula to calculate the heat lost (Q) by the copper block when it cools down is given by:

\[ Q = m \cdot s \cdot \Delta T \]

Where:
- \( m \) = mass of the copper block = 2 kg
- \( s \) = specific heat capacity of copper = 400 J/kg/°C
- \( \Delta T \) = change in temperature = initial temperature - final temperature = 500°C - 0°C = 500°C

Substituting the values:

\[ Q = 2 \, \text{kg} \cdot 400 \, \text{J/kg/°C} \cdot 500 \, \text{°C} \]

Calculating this gives:

\[ Q = 2 \cdot 400 \cdot 500 = 400000 \, \text{J} \]

### Step 2: Relate the heat gained by the ice to the heat lost by the copper block

The heat gained by the ice (which melts) can be expressed as:

\[ Q = m_{\text{ice}} \cdot L_f \]

Where:
- \( m_{\text{ice}} \) = mass of ice melted (in kg)
- \( L_f \) = latent heat of fusion of water = \( 3.5 \times 10^5 \, \text{J/kg} \)

### Step 3: Set the heat lost equal to the heat gained

Since the heat lost by the copper block is equal to the heat gained by the ice, we can write:

\[ 400000 \, \text{J} = m_{\text{ice}} \cdot 3.5 \times 10^5 \, \text{J/kg} \]

### Step 4: Solve for the mass of ice melted

Rearranging the equation to find \( m_{\text{ice}} \):

\[ m_{\text{ice}} = \frac{400000 \, \text{J}}{3.5 \times 10^5 \, \text{J/kg}} \]

Calculating this gives:

\[ m_{\text{ice}} = \frac{400000}{350000} = \frac{8}{7} \, \text{kg} \]

### Conclusion

The amount of ice that can melt is:

\[ m_{\text{ice}} = \frac{8}{7} \, \text{kg} \approx 1.14 \, \text{kg} \]