Let's solve the problem step by step. --- ## **Given Data** - Mass of water (\( m \)) = **1 kg** - Initial temperature (\( T_1 \)) = **250 K** - Final temperature (\( T_5 \)) = **400 K** - Standard atmospheric pressure. - Heat capacity of ice (250 K to 273 K) = **2.037 kJ/kg·K** - Mean heat capacity of liquid water (273 K to 373 K) = **75.726 kJ/kmol·K** - Heat capacity of water vapor (\( C \), kJ/kmol·K): \( C = 30.475 + 9.652 \times 10^{-3} T + 1.189 \times 10^{-7} T^2 \) - Latent heat of fusion = **6012 kJ/kmol** - Latent heat of vaporization = **40608 kJ/kmol** - Molar mass of water (\( M \)) = **18 kg/kmol** --- ## **Step 1: Calculate the heat required in different steps** ### **Step 1: Heating ice from 250 K to 273 K** \[ Q_1 = m \cdot c_{ice} \cdot (273 - 250) \] \[ Q_1 = 1 \cdot 2.037 \cdot (273 - 250) \] \[ Q_1 = 2.037 \cdot 23 = \boxed{46.85 \text{ kJ}} \] --- ### **Step 2: Melting ice at 273 K (fusion)** \[ Q_2 = n \cdot \Delta H_{fus} \] where \( n = \frac{m}{M} = \frac{1}{18} \) kmol \[ Q_2 = \frac{1}{18} \cdot 6012 = \boxed{334 \text{ kJ}} \] --- ### **Step 3: Heating water from 273 K to 373 K** Use mean heat capacity of 75.726 kJ/kmol·K. \[ Q_3 = n \cdot c_{water} \cdot (373 - 273) \] \[ Q_3 = \frac{1}{18} \cdot 75.726 \cdot 100 \] \[ Q_3 = .05556 \cdot 7572.6 = \boxed{420.7 \text{ kJ}} \] --- ### **Step 4: Vaporizing water at 373 K** \[ Q_4 = n \cdot \Delta H_{vap} \] \[ Q_4 = \frac{1}{18} \cdot 40608 = \boxed{2256 \text{ kJ}} \] --- ### **Step 5: Heating vapor from 373 K to 400 K** \[ Q_5 = n \int_{373}^{400} C_{vapor} dT \] Where \( C_{vapor} = 30.475 + 9.652 \times 10^{-3}T + 1.189 \times 10^{-7}T^2 \) Let’s integrate: \[ \int_{373}^{400} \left( 30.475 + 9.652 \times 10^{-3}T + 1.189 \times 10^{-7}T^2 \right) dT \] \[ = 30.475 (T_2 - T_1) + \frac{9.652 \times 10^{-3}}{2}(T_2^2 - T_1^2) + \frac{1.189 \times 10^{-7}}{3}(T_2^3 - T_1^3) \] Plugging in \( T_2 = 400 \), \( T_1 = 373 \): #### 1. \( 30.475 \times (400-373) = 30.475 \times 27 = 822.825 \) #### 2. \( T_2^2 - T_1^2 = 400^2 - 373^2 = 160000 - 139129 = 20871 \) \[ \frac{9.652 \times 10^{-3}}{2} \times 20871 = .004826 \times 20871 = 100.7 \] #### 3. \( T_2^3 - T_1^3 = 64,000,000 - 51,876,317 = 12,123,683 \) \[ \frac{1.189 \times 10^{-7}}{3} \times 12,123,683 = 3.963 \times 10^{-8} \times 12,123,683 = .4806 \] Add these: \[ 822.825 + 100.7 + .4806 = 924.01 \text{ kJ/kmol} \] Now, multiply by \( n \): \[ Q_5 = \frac{1}{18} \times 924.01 = \boxed{51.33 \text{ kJ}} \] --- ## **Step 6: Total Heat Required** \[ Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 \] \[ Q_{total} = 46.85 + 334 + 420.7 + 2256 + 51.33 \] \[ Q_{total} = \boxed{3108.88 \text{ kJ}} \] --- ## **Final Answer** \[ \boxed{3109 \ \text{ kJ}} \] (rounding to 4 significant digits) --- ### **Summary Table** | Step | Description | Heat (kJ) | |------|-------------------------------------|-----------| | 1 | Ice: 250 K → 273 K | 46.85 | | 2 | Melt ice (fusion) | 334 | | 3 | Water: 273 K → 373 K | 420.7 | | 4 | Vaporize at 373 K | 2256 | | 5 | Vapor: 373 K → 400 K | 51.33 | | | **Total** | **3109** | --- **The heat required to heat 1 kg of water from 250 K to 400 K at 1 atm is approximately \( \boxed{3109 \text{ kJ}} \).**