Let's break down the problem and solve it step by step: ## **Given Data** - **Feed rate:** 300 kg/h - **Feed temperature:** 367 K - **Final temperature:** 333 K - **MgSO₄·6H₂O in feed:** 45 kg/100 kg water - **Heat capacity of solution (\(C_p\)):** 3.2 kJ/kg·K - **Mother liquor output:** 180 kg/h - **No water evaporated** - **Find:** Total heat load (kW) --- ## **Step 1: Amount of crystals removed per hour** ### **Material Balance** Let: - \( F = \) Feed rate = 300 kg/h - \( M = \) Mother liquor = 180 kg/h - \( C = \) Crystals formed = ? \[ C = F - M = 300 - 180 = \boxed{120~\text{kg/h}} \] --- ## **Step 2: Mass of MgSO₄·6H₂O in crystals** All the crystals removed are MgSO₄·6H₂O: \[ \text{Mass of crystals (MgSO}_4\cdot 6\text{H}_2\text{O)} = 120~\text{kg/h} \] --- ## **Step 3: Calculate Heat Removed (Q)** ### **Heat removed = Heat removed from cooling + Heat of crystallization** #### **A. Cooling the feed from 367 K to 333 K** \[ Q_1 = \text{mass cooled} \times C_p \times \Delta T \] All 300 kg/h is cooled: \[ Q_1 = 300~\text{kg/h} \times 3.2~\text{kJ/kg.K} \times (367-333)~\text{K} \] \[ Q_1 = 300 \times 3.2 \times 34 \] \[ Q_1 = 300 \times 108.8 \] \[ Q_1 = 326400~\text{kJ/h} \] Convert to kW (\(1~\text{kW} = 360~\text{kJ/h}\)): \[ Q_1 = \dfrac{326400}{360} = 90.67~\text{kW} \] --- #### **B. Heat of crystallization** This information is **not given** in the problem. So, we **ignore** it as per instructions. --- ## **Step 4: Final Answer** The closest option to 90.67 kW is **94 kW**. ### **Final Answer** \[ \boxed{94~\text{kW}} \] --- ### **Summary Table** | Step | Calculation | Value | |----------------|----------------------|-----------| | Crystals | 300 - 180 | 120 kg/h | | Heat load | \(Q = mC_p\Delta T\) | 94 kW | --- ### **Answer:** **94 kW**