​A copper block of mass 3 kg is heated in a furnace to a temperature of 450 °C and then placed on a large ice block. Find the maximum amount of ice that can melt? (specific heat of copper = 0.39 $$\text{Jg}^{-1}\text{K}^{-1}$$​, heat of fusion of water = 335 $$\text{Jg}^{-1}\text{K}^{-1}$$​)​

Correct option is B

​$$\text{By the principle of calorimetry:} \\\text{Heat given by hot body} = \text{Heat gained by cold body} \\\text{In this case, the heat given by the copper block is equal to the heat gained by the ice block:} \\m_{\text{Cu}} c_{\text{Cu}} (T_{\text{Cu}} - T_{\text{final}}) = m_{\text{ice}} L_f \\\text{Where:} \\m_{\text{Cu}} \text{ is the mass of the copper block,} \\c_{\text{Cu}} \text{ is the specific heat of copper,} \\T_{\text{Cu}} \text{ is the initial temperature of the copper block,} \\T_{\text{final}} \text{ is the final temperature,} \\m_{\text{ice}} \text{ is the mass of ice melted,} \\L_f \text{ is the latent heat of fusion of water.}$$​

$$\text{Given:} \\m_{\text{Cu}} = 3 \, \text{kg}, \\c_{\text{Cu}} = 0.39 \, \text{J/g}^\circ \text{C}, \\T_{\text{Cu}} = 450^\circ \text{C}, \\L_f = 335 \, \text{J/g}. \\\text{Substituting the values into the formula:} \\m_{\text{ice}} = \dfrac{m_{\text{Cu}} c_{\text{Cu}} (T_{\text{Cu}} - T_{\text{final}})}{L_f} \\\text{Since } T_{\text{final}} \text{ is } 0^\circ \text{C (for melting ice):} \\m_{\text{ice}} = \dfrac{3 \times 0.39 \times (450 - 0)}{335} \\m_{\text{ice}} \approx 1.57 \, \text{kg}$$​​