Solve correct option: a
Question 4-0 Max-score- LOU
= veigeor operating between tempeceure ts of 0° C 40d 30°C, removing 50 J of energy from lower temperature end rejecting 60 kJ of heat to the ighe

Question

Solve correct option: a
Question 4-0 Max-score- LOU
= veigeor operating between tempeceure ts of 0° C 40d 30°C, removing 50 J of energy from lower temperature end rejecting 60 kJ of heat to the igher
Temperature, Caicusie the exergy efficiency of the cycle.

VIPSolutions · Accepted Answer

Let's solve the problem step by step.

### **Given Data**
- Refrigerator operates between $ T_L = ^\circ C = 273 \ K and \( T_H = 30^\circ C = 303 $ K.
- $ Q_L = 50 $ kJ (heat removed from low temperature)
- $ Q_H = 60 $ kJ (heat rejected to high temperature)

### **Exergy Efficiency (Second Law Efficiency)**
The exergy efficiency for a refrigerator is:
$$
\text{Exergy efficiency} = \frac{\text{Minimum work required}}{\text{Actual work input}}
$$

#### **Step 1: Actual Work Input**
$$
W_{\text{actual}} = Q_H - Q_L = 60\,\text{kJ} - 50\,\text{kJ} = 10\,\text{kJ}
$$

#### **Step 2: Minimum Work Required**
The minimum (reversible) work input for a refrigerator is:
$$
W_{\text{min}} = Q_L \left( \frac{T_H - T_L}{T_L} \right)
$$
$$
= 50 \left( \frac{303 - 273}{273} \right)
$$
$$
= 50 \left( \frac{30}{273} \right)
$$
$$
= 50 \times .10989 \approx 5.4945\,\text{kJ}
$$

#### **Step 3: Exergy Efficiency Calculation**
$$
\text{Exergy efficiency} = \frac{W_{\text{min}}}{W_{\text{actual}}}
$$
$$
= \frac{5.4945}{10} = .549
$$

---

## **Final Answer**
**The correct option is:**

$$
\boxed{.549}
$$

✨ VIPSolutions

Solve correct option: a Question 4-0 Max-score- LOU = veigeor operating between tempeceure ts of 0° C 40d 30°C, removing 50 J of energy from lower temperature end rejecting 60 kJ of heat to the igher Temperature, Caicusie the exergy efficiency of the cycle.

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