A copper block of mass `2.5 kg` is heated in a furnace to a temperature of `500^(@)C` and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat copper `= 0.39 J//g^(@)C` heat of fusion of water `= 335 J//g`).

Options

A. `2 kg`
B. `1.5 kg`
C. `2.5 kg`
D. `0.5 kg`

Explanation

To solve the problem of how much ice can be melted by a copper block heated to 500°C, we can follow these steps:

### Step 1: Calculate the heat lost by the copper block
The heat lost by the copper block when it cools down to 0°C can be calculated using the formula:

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

Where:
- \( Q \) = heat lost (in joules)
- \( m \) = mass of the copper block (in grams)
- \( c \) = specific heat capacity of copper (in J/g°C)
- \( \Delta T \) = change in temperature (in °C)

Given:
- Mass of copper block, \( m = 2.5 \, \text{kg} = 2500 \, \text{g} \)
- Specific heat of copper, \( c = 0.39 \, \text{J/g°C} \)
- Initial temperature of copper, \( T_i = 500°C \)
- Final temperature of copper, \( T_f = 0°C \)

The change in temperature, \( \Delta T = T_i - T_f = 500 - 0 = 500°C \).

Now substituting the values into the formula:

\[ Q = 2500 \, \text{g} \cdot 0.39 \, \text{J/g°C} \cdot 500 \, °C \]

Calculating this gives:

\[ Q = 2500 \cdot 0.39 \cdot 500 = 487500 \, \text{J} \]

### Step 2: Calculate the mass of ice that can be melted
The heat required to melt ice can be calculated using the formula:

\[ Q = m \cdot L \]

Where:
- \( Q \) = heat absorbed by the ice (in joules)
- \( m \) = mass of ice melted (in grams)
- \( L \) = latent heat of fusion of ice (in J/g)

Given:
- Latent heat of fusion of ice, \( L = 335 \, \text{J/g} \)

Now we can rearrange the formula to find the mass of ice melted:

\[ m = \frac{Q}{L} \]

Substituting the values we have:

\[ m = \frac{487500 \, \text{J}}{335 \, \text{J/g}} \]

Calculating this gives:

\[ m \approx 1455.22 \, \text{g} \]

### Step 3: Convert grams to kilograms
To express the mass in kilograms:

\[ m \approx 1.455 \, \text{kg} \]

### Conclusion
The maximum amount of ice that can melt is approximately **1.455 kg**, which can be rounded to **1.5 kg**.