Let's break down the problem step-by-step: **Given:** - Initial temperature of iron, \( T_{i,Fe} = 225.0^\circ C \) - Final temperature of iron, \( T_{f,Fe} = 0.0^\circ C \) - Mass of ice, \( m_{ice} = 1.21 \) g - Initial temperature of ice, \( T_{i,ice} = -5.00^\circ C \) - Final temperature of ice (after melting), \( T_{f,ice} = 0.0^\circ C \) - Find: **Mass of iron (\( m_{Fe} \))** **Heat lost by iron = Heat gained by ice** ### Step 1: Calculate Heat Gained by Ice The ice at \(-5.00^\circ C\) must: 1. Warm up to \(0^\circ C\) 2. Melt at \(0^\circ C\) **a) Warm ice from -5.00°C to 0.00°C** \[ q_1 = m_{ice} \cdot c_{ice} \cdot \Delta T \] Where: - \(c_{ice} = 2.09 \, \text{J/g}^\circ\text{C}\) (specific heat of ice) - \(\Delta T = 0.00 - (-5.00) = 5.00^\circ C\) \[ q_1 = 1.21 \, \text{g} \times 2.09 \, \text{J/g}^\circ\text{C} \times 5.00^\circ C = 12.64 \, \text{J} \] **b) Melt ice at 0.00°C** \[ q_2 = m_{ice} \cdot \Delta H_f \] Where: - \(\Delta H_f = 334 \, \text{J/g}\) (heat of fusion for ice) \[ q_2 = 1.21 \, \text{g} \times 334 \, \text{J/g} = 404.14 \, \text{J} \] **Total heat absorbed by ice:** \[ q_{ice,total} = q_1 + q_2 = 12.64 \, \text{J} + 404.14 \, \text{J} = 416.78 \, \text{J} \] --- ### Step 2: Calculate Heat Lost by Iron \[ q_{Fe} = m_{Fe} \cdot c_{Fe} \cdot \Delta T \] Where: - \(c_{Fe} = 0.449 \, \text{J/g}^\circ\text{C}\) (specific heat of iron) - \(\Delta T = 225.0^\circ C - 0.0^\circ C = 225.0^\circ C\) --- ### Step 3: Set Heat Lost by Iron Equal to Heat Gained by Ice \[ q_{Fe} = q_{ice,total} \] \[ m_{Fe} \cdot 0.449 \cdot 225.0 = 416.78 \] \[ m_{Fe} = \frac{416.78}{0.449 \cdot 225.0} \] \[ m_{Fe} = \frac{416.78}{101.025} \] \[ m_{Fe} \approx 4.13 \, \text{g} \] --- ## **Final Answer** \[ \boxed{4.13 \text{ g}} \] **So the mass of the piece of iron is 4.13 grams.**

The heat lost by the iron as it cools from 225°C to 0°C equals the heat absorbed by the ice to warm from -5°C to 0°C and melt at 0°C. Using specific heats and latent heat, the iron's mass is calculated as approximately 4.13 grams.