Let's break down the steps to convert \(0.500 \, \text{kg}\) of ice at \(-20^\circ \text{C}\) to steam at \(250^\circ \text{C}\): **Step 1:** Heat ice from \(-20^\circ \text{C}\) to \(0^\circ \text{C}\). - \(q_1 = m c_{\text{ice}} \Delta T\) - \(c_{\text{ice}} = 2.1 \, \text{J/g}^\circ\text{C}\) - \(\Delta T = 0 - (-20) = 20^\circ \text{C}\) - \(m = 0.500\, \text{kg} = 500\, \text{g}\) \[ q_1 = 500 \times 2.1 \times 20 = 21,000 \, \text{J} \] **Step 2:** Melt ice at \(0^\circ \text{C}\) to water at \(0^\circ \text{C}\). - \(q_2 = m L_f\) - \(L_f = 334 \, \text{J/g}\) \[ q_2 = 500 \times 334 = 167,000 \, \text{J} \] **Step 3:** Heat water from \(0^\circ \text{C}\) to \(100^\circ \text{C}\). - \(q_3 = m c_{\text{water}} \Delta T\) - \(c_{\text{water}} = 4.18 \, \text{J/g}^\circ\text{C}\) - \(\Delta T = 100 - 0 = 100^\circ \text{C}\) \[ q_3 = 500 \times 4.18 \times 100 = 209,000 \, \text{J} \] **Step 4:** Vaporize water at \(100^\circ \text{C}\) to steam at \(100^\circ \text{C}\). - \(q_4 = m L_v\) - \(L_v = 2260 \, \text{J/g}\) \[ q_4 = 500 \times 2260 = 1,130,000 \, \text{J} \] **Step 5:** Heat steam from \(100^\circ \text{C}\) to \(250^\circ \text{C}\). - \(q_5 = m c_{\text{steam}} \Delta T\) - \(c_{\text{steam}} = 2.0 \, \text{J/g}^\circ\text{C}\) - \(\Delta T = 250 - 100 = 150^\circ \text{C}\) \[ q_5 = 500 \times 2.0 \times 150 = 150,000 \, \text{J} \] **Total Energy:** \[ q_{\text{total}} = q_1 + q_2 + q_3 + q_4 + q_5 = 21,000 + 167,000 + 209,000 + 1,130,000 + 150,000 = 1,677,000 \, \text{J} \] Convert to kJ: \[ 1,677,000 \, \text{J} = 1,677 \, \text{kJ} \approx 1680 \, \text{kJ} \] **Answer:** **d) 1680 kJ**