Let's solve the heat exchanger problem step by step, as this is the main question with relevant data: ## **Given Data** - Hot oil enters at \(200^\circ \text{C}\) - Hot oil leaves at \(150^\circ \text{C}\) - Mass flow rate of oil (\(m_{hot}\)) = 2 kg/s - Water enters at \(30^\circ \text{C}\) - Mass flow rate of water (\(m_{cold}\)) = 2 kg/s - Specific heat of oil (\(cp_{hot}\)) = 2.5 kJ/kg·K - Specific heat of water (\(cp_{cold}\)) = 4.18 kJ/kg·K ## **Step 1: Calculate Heat Lost by Oil (\(Q_{hot}\))** \[ Q = m \cdot c_p \cdot \Delta T \] \[ Q_{hot} = 2 \text{ kg/s} \times 2.5 \text{ kJ/kg·K} \times (200 - 150)^\circ \text{C} \] \[ Q_{hot} = 2 \times 2.5 \times 50 \] \[ Q_{hot} = 5 \times 50 = 250 \text{ kJ/s} = 250 \text{ kW} \] ## **Step 2: Calculate Water Outlet Temperature (\(T_{water, out}\))** The heat gained by water = heat lost by oil (assuming no heat loss to surroundings): \[ Q_{cold} = m_{cold} \cdot c_{p,cold} \cdot (T_{water, out} - T_{water, in}) \] \[ 250 = 2 \times 4.18 \times (T_{water, out} - 30) \] \[ 250 = 8.36 \times (T_{water, out} - 30) \] \[ T_{water, out} - 30 = \frac{250}{8.36} \approx 29.9 \] \[ T_{water, out} = 29.9 + 30 = 59.9^\circ \text{C} \] ## **Step 3: Summary of Answers** - **Outlet temperature of water:** \(59.9^\circ \text{C}\) - **Heat transferred:** \(250 \text{ kW}\) --- ### **Final Answers** 1. **Heat transferred (\(Q\))**: \(250 \text{ kW}\) 2. **Outlet temperature of water**: \(59.9^\circ \text{C}\) --- If you need the calculation for any other part (e.g., log mean temperature difference or area), please specify!