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. The maximum amount of ice that can be melt, will be :
Given :
Specific heat of copper = 0.4 J g$^{-1}$°C$^{-1}$
Heat of fusion of water = 400 Jg$^{-1}$

Question

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. The maximum amount of ice that can be melt, will be : 
Given : 
Specific heat of copper = 0.4 J g$^{-1}$°C$^{-1}$ 
Heat of fusion of water = 400 Jg$^{-1}$

Accepted Answer

To determine the maximum amount of ice that can be melted by the copper block, we need to calculate the amount of heat transferred from the copper block to the ice. This heat will be used to melt the ice.

1. **Calculate the heat lost by the copper block:**

The formula to calculate the heat lost by the copper block is:
   $$
   Q = m \cdot c \cdot \Delta T
   $$
   where:
   - $ Q $ is the heat lost,
   - $ m $ is the mass of the copper block,
   - $ c $ is the specific heat capacity of copper,
   - $ \Delta T $ is the change in temperature.

Given:
   - Mass of copper block, $ m = 2.5 \, \text{kg} = 2500 \, \text{g} $ (since 1 kg = 1000 g),
   - Specific heat of copper, $ c = 0.4 \, \text{J g}^{-1}°\text{C}^{-1} $,
   - Initial temperature of copper, $ T_i = 500°C $,
   - Final temperature of copper, $ T_f = 0°C $ (since it is placed on ice).

$$
   \Delta T = T_i - T_f = 500°C - 0°C = 500°C
   $$

Substituting the values:
   $$
   Q = 2500 \, \text{g} \times 0.4 \, \text{J g}^{-1}°\text{C}^{-1} \times 500°C = 500,000 \, \text{J}
   $$

2. **Calculate the mass of ice melted:**

The heat required to melt ice is given by:
   $$
   Q = m_{\text{ice}} \cdot L_f
   $$
   where:
   - $ m_{\text{ice}} $ is the mass of ice melted,
   - $ L_f $ is the heat of fusion of water.

Given:
   - Heat of fusion of water, $ L_f = 400 \, \text{J g}^{-1} $.

Rearranging the formula to solve for $ m_{\text{ice}} $:
   $$
   m_{\text{ice}} = \frac{Q}{L_f} = \frac{500,000 \, \text{J}}{400 \, \text{J g}^{-1}} = 1250 \, \text{g}
   $$

Converting grams to kilograms:
   $$
   m_{\text{ice}} = 1250 \, \text{g} = 1.25 \, \text{kg}
   $$

Therefore, the maximum amount of ice that can be melted is 1.25 kg, which corresponds to **Option 1**.

**Explanation for Incorrect Options:**

- **Option 2 (1.50 kg):** This option suggests that more ice is melted than the calculated 1.25 kg. However, based on the heat transfer calculations, the energy available from the copper block is only sufficient to melt 1.25 kg of ice.

- **Option 3 (1.75 kg):** This option also overestimates the amount of ice melted. The calculations clearly show that the heat from the copper block can only melt 1.25 kg of ice.

- **Option 4 (1.80 kg):** This option further overestimates the ice melting capacity. The energy provided by the copper block is not enough to melt 1.80 kg of ice, as demonstrated by the calculations.

In conclusion, the calculations confirm that **Option 1 (1.25 kg)** is the correct answer, as it accurately reflects the maximum amount of ice that can be melted by the heat lost from the copper block.