A metal block of mass \(3.3\,\text{kg}\) is heated to a temperature of \(400^\circ\text{C}\) and then block is placed on a large ice block. The specific heat of the metal is \(0.4\,\text{J g}^{-1}\text{K}^{-1}\) and heat of fusion of water is \(330\,\text{J g}^{-1}\). The maximum amount of ice that can melt is
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For ice-melting problems, equate heat lost by the hot body to latent heat gained by ice:
\[
mc\Delta T=m_{\text{ice}}L
\]
Step 1: Understand the heat transfer process. When a hot metal block is placed on an ice block, heat flows from the metal to the ice. The metal cools from $400^\circ\text{C}$ to $0^\circ\text{C}$ (the melting point of ice), and this released heat is used to melt ice. We assume no heat loss to surroundings, which gives us the maximum amount of ice that can melt. Step 2: Convert mass of metal to grams. The mass given is $3.3\,\text{kg}$. Since the specific heat is in $\text{J g}^{-1}\text{K}^{-1}$, we convert: \[ m_{\text{metal}} = 3.3\,\text{kg} = 3300\,\text{g} \] Step 3: Calculate heat lost by the metal. The heat lost by the metal as it cools from $400^\circ\text{C}$ to $0^\circ\text{C}$ is given by: \[ Q = m \cdot c \cdot \Delta T \] where $c = 0.4\,\text{J g}^{-1}\text{K}^{-1}$ and $\Delta T = 400\,\text{K}$. So: \[ Q = 3300 \times 0.4 \times 400 = 528000\,\text{J} \] Step 4: Apply conservation of energy for ice melting. All the heat released by the metal goes into melting ice. The heat required to melt a mass $m_{\text{ice}}$ of ice is: \[ Q = m_{\text{ice}} \times L \] where $L = 330\,\text{J g}^{-1}$ is the latent heat of fusion. So we write: \[ m_{\text{ice}} = \frac{Q}{L} \] Step 5: Compute the mass of ice melted. Substituting the values: \[ m_{\text{ice}} = \frac{528000}{330} = 1600\,\text{g} \] Converting to kilograms: \[ m_{\text{ice}} = 1.6\,\text{kg} \] Step 6: State the final answer. The maximum amount of ice that can melt is: \[ \boxed{1.6\,\text{kg}} \]
