Question

One hundred kg of wet food contains 60 per cent moisture content on wet basis is to be dried to 20 per cent moisture content on wet basis. What is the weight of water evaporated?

• A. 25 kg
• B. 50 kg
• C. 75 kg
• D. 100 kg

Hint:

In drying problems, dry solids remain constant. Always use this principle.

Answer 1

The Correct Option is B

Concept: Moisture content on wet basis is defined as: \[ \text{Moisture content (wb)} = \frac{\text{Weight of water}}{\text{Total weight of material}} \times 100 \] During drying, the dry matter remains constant while water content decreases.

Step 1: Initial composition.
Total initial weight = 100 kg
Moisture content = 60% (wet basis) \[ \text{Initial water} = 0.60 \times 100 = 60 \text{ kg} \] \[ \text{Dry solids} = 100 - 60 = 40 \text{ kg} \]

Step 2: Final condition.
Final moisture content = 20% (wet basis) Let final total weight = $W$ \[ \text{Water} = 0.20W \] \[ \text{Dry solids} = 0.80W \]

Step 3: Using constant dry solids condition.
Dry solids remain unchanged: \[ 0.80W = 40 \] \[ W = \frac{40}{0.80} = 50 \text{ kg} \]

Step 4: Final water content. \[ \text{Final water} = 0.20 \times 50 = 10 \text{ kg} \]

Step 5: Water evaporated. \[ \text{Water removed} = 60 - 10 = 50 \text{ kg} \]

Step 6: Matching options.
Option (2) matches the result. Final Conclusion:
The amount of water evaporated is 50 kg. Hence, the correct answer is option (2).