Question

1000 kg of grain at 10% (wet basis) moisture was stored in coastal region for six months, where grain absorbed 50 kg water vapour. The amount of dry matter content in the final product will be

• A. 900 kg
• B. 1000 kg
• C. 950 kg
• D. 1050 kg

Hint:

Dry matter is indestructible during sorption and drying! Only the moisture content varies: $1000 \times (1 - 0.10) = 900\text{ kg}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In post-harvest grain storage and drying processes, the total mass of bone-dry matter remains strictly constant across all moisture gain or loss operations (Law of Conservation of Mass).
Key Formula or Approach:
\[ \text{Dry Matter Mass } (M_{dm}) = M_{\text{total}} \times \left( 1 - \frac{M_{wb}}{100} \right) \]

Step 2: Detailed Explanation:
Given data:
- Initial total mass of grain: \(M_1 = 1000\text{ kg}\)
- Initial moisture content (wet basis): \(M_{wb} = 10\% = 0.10\)
Initial water mass:
\[ M_{\text{water}} = 1000\text{ kg} \times 0.10 = 100\text{ kg} \]
Initial dry matter mass:
\[ M_{dm} = 1000\text{ kg} - 100\text{ kg} = 900\text{ kg} \]
When the grain absorbs 50 kg of ambient water vapour, the new total mass becomes \(1050\text{ kg}\) and total water becomes \(150\text{ kg}\).
The dry matter content undergoes no chemical destruction or addition, remaining identically equal to \(900\text{ kg}\).

Step 3: Final Answer:
Therefore, the amount of dry matter content in the final product is 900 kg, matching option (A).