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

Answer

To solve the problem, we need to calculate the amount of ice that can melt when a 2 kg copper block at 500°C is placed on it. We will follow these steps:

### 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:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( m \) = mass of the copper block = 2 kg
- \( c \) = 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:
\[ Q = 2 \, \text{kg} \cdot 400 \, \text{J/kg/°C} \cdot 500 \, °C \]
\[ Q = 2 \cdot 400 \cdot 500 = 400000 \, \text{J} \]
So, the heat lost by the copper block is \( 4 \times 10^5 \, \text{J} \).

### Step 2: Calculate the mass of ice that can be melted
The heat required to melt ice is given by:
\[ Q = m \cdot L_f \]
where:
- \( m \) = mass of ice melted (in kg)
- \( L_f \) = latent heat of fusion of ice = \( 3.5 \times 10^5 \, \text{J/kg} \)

We can rearrange the equation to find the mass of ice melted:
\[ m = \frac{Q}{L_f} \]

Substituting the values we have:
\[ m = \frac{400000 \, \text{J}}{3.5 \times 10^5 \, \text{J/kg}} \]
\[ m = \frac{4 \times 10^5}{3.5 \times 10^5} \]
\[ m = \frac{4}{3.5} \]
\[ m = \frac{40}{35} = \frac{8}{7} \]

### Final Answer
The amount of ice that can melt is \( \frac{8}{7} \, \text{kg} \).