# Step-by-Step Solution: Finding COP of a Refrigerator Let's break down the problem and solve it step by step. --- ## **Given** - **Cooling effect (Heat removed from the cold space, \( Q_L \))**: 420 kJ/min - **Work input (\( W \))**: 140 kJ/min --- ## **Step 1: Understanding COP** The **Coefficient of Performance (COP)** for a refrigerator is defined as: \[ COP_{ref} = \frac{\text{Desired effect (Cooling)}}{\text{Work input}} \] So, \[ COP_{ref} = \frac{Q_L}{W} \] where: - \( Q_L \) = heat removed from the cold region (in kJ/min) - \( W \) = work input (in kJ/min) --- ## **Step 2: Substitute Values** Given: - \( Q_L = 420 \) kJ/min - \( W = 140 \) kJ/min Substitute these values into the formula: \[ COP_{ref} = \frac{420}{140} \] --- ## **Step 3: Calculate** \[ \frac{420}{140} = 3 \] --- ## **Step 4: Write the Final Answer** \[ \boxed{COP_{ref} = 3} \] --- ## **Explanation** - The COP tells us how many units of heat are removed from the refrigerator for each unit of work supplied. - Here, for every 1 kJ of work, the refrigerator removes 3 kJ of heat from the cold space. - A higher COP means a more efficient refrigerator. --- **Summary Table** | Quantity | Value | Units | |---------------|--------------|--------| | Cooling (\(Q_L\)) | 420 kJ/min | kJ/min | | Work Input (\(W\))| 140 kJ/min | kJ/min | | COP | 3 | – | --- **Final Answer:** The COP of the refrigerator is **3**.