# Fruits Processing Questions and Answers – Rate of Dehydration – Numericals

This set of Fruits Processing Multiple Choice Questions & Answers (MCQs) focuses on “Rate of Dehydration – Numericals”.

1. A dryer is used for drying pears which has initial moisture content of 70 % (wet basis) and final moisture content of 5 % (wet basis). An experimental drying curve for the product indicates that the critical moisture content is 25 % (wet basis) and the time for constant drying is 5 min. Can you estimate the total drying time for product?
a) 6.54 min
b) 64 min
c) 654 min
d) 4.56 min

Explanation:
Initial product moisture content (Wo) = 0.7 / 0.3 = 2.33 kg (water) / kg (solids)
Critical moisture content (Wc) = 0.25 / 0.75 = 0.333 kg (water) / kg (solids)
Final moisture content (Wf) = 0.05 / 0.95 = 0.0526 kg (water) / kg (solids)
Time for constant rate drying (tc) = 5 min
Required Total drying time (Rc) = ((Wo) – (Wc)) / tc
= (2.33 – 0.33) / 5 min
= 0.4 kg (water) / kg (solids) min
Falling rate drying time (tF) = (Wc)/Rc * ln ((Wc)/w)
= 0.333/0.4 * ln (0.333 / 0.0526)
= 1.54 min
Total drying time (t) = 5 + 1.54 = 6.54 min

2. An apple contains 70% of water which has to be dried at 100°C down to moisture content of 15%. If the initial temperature of the food is 21°C, The latent heat of vaporization of water at 100°C and at standard atmospheric pressure is 2257kJkg-1. The specific heat capacity of the food is 3.8 kJkg-1°C-1 and of water is 4.186 kJkg-1°C-1. Calculate the quantity of heat energy required per unit weight of the original material, for drying under atmospheric pressure?
a) 1810kJ
b) 1096kJ
c) 2199kJ
d) 2032kJ

Explanation:
Calculating for 1 kg food,
Initial moisture = 70%
700g moisture is associated with 300 g dry matter
Final moisture = 15 %
15g moisture is associated with 985 g dry matter
Therefore (100 * 300)/900g = 30.6g moisture are associated with 300 g dry matter.
1kg of original matter must lose (700 -30.6) g moisture = 669.4 g
= 0.669 kg moisture
Heat energy required for 1kg original material
= heat energy to raise temperature to 100°C+ latent heat to remove water
= (100 -21) * 3.8 + 0.669 *2257
= 300.2 + 1509.9
= 1810.1kJ

3. If the weight of the hot yellow pepper cubes is 170.2grams and contains 165.5 grams of moisture content, what is the wet basis moisture content of the pepper?
a) 99%
b) 97.2%
c) 89.3%
d) 93.3%

Explanation:
% Wet Basis Moisture = Weight of Water in the Sample * 100 / %Total Weight of the Sample
= 165.5 * 100/ 170.2
= 97.2%

4. If 232.9 grams of grapes contains 198 grams of water, what is the dry basis moisture content of the grapes?
a) 5.67%
b) 6.9%
c) 8.75%
d) 9.9%

Explanation:
Total Weight of Material = Weight of Water + Weight of Dry Solids
Therefore, Weight of Dry Solids = Total Weight of Material- Weight of Water
= 232.9 – 198
= 34.9 grams of dry solids
Dry Basis Moisture % = Weight of Water in the Sample * 100 / Weight of Solids in the Sample
= 198 * 100 / 34.9
= 5.67 %

5. The eggplant has a dry basis moisture content of 8.9 grams of water per gram of dry solids. What is its wet basis moisture content?
a) 85.8%
b) 88.0%
c) 89.8%
d) 90.8%

Explanation:
% Wet Basis Moisture = Weight of Water in the Sample * 100 / % of Total Weight of the Sample
Total weight of the sample= 8.9 g of water + 1.00 g of dry solids.
= 9.9g
% Wet Basis Moisture = 8.9 * 100 / 9.9
= 89.8%

6. A processor has 100 kg of mangoes with a moisture content of 85% on a wet basis. Calculate the weight of water and solids present respectively.
a) 80kg and 15kg
b) 85kg and 15kg
c) 95kg and 25kg
d) 95kg and 15kg

Explanation:
Weight of water = Weight of material * percentage water as a decimal
= 100 kg * 0.85
= 85 kg
% Solids = 100% – % Water
= 100% – 85% = 15%
Weight of solids = Weight of material * percent solids as a decimal
= 100 kg * 0.15 = 15kg
Therefore, there are 85kg of water and 15 kg of solids (i.e., dry solids) present.

7. If 456.7 grams of mushroom contains 400 grams of water, what is the wet basis moisture content of the mushroom?
a) 99%
b) 97.2%
c) 87.5%
d) 93.3%

Explanation:
% Wet Basis Moisture = Weight of Water in the Sample * 100 / %Total Weight of the Sample
= 400 * 100/ 456.7
= 97.2%

8. How many kilograms of cherry tomatoes are required to obtain 150 kg of dried cherry tomatoes with a final moisture content of 11% wet basis? The ripe cherry tomatoes used have a moisture content of 94% wet basis.
a) 225 kg
b) 3225 kg
c) 3925 kg
d) 2225 kg

Explanation:
Weight of water = Weight of material * percentage water as a decimal
Weight of water in cherry tomatoes = 150 kg * 0.11 (i.e., 11% as a decimal)
= 16.5 kg
% solids in cherry tomatoes = 100% – % water in product
= 100% -11%= 89%
Weight of solids in cherry tomatoes = 150 kg * 0.89 (i.e., 89% as a decimal)
= 133.5 kg
Since there is no indication that any solids were lost in the drying process, we assume that the weight of solids present in the starting material is the same as that present in the finished product.
Weight of solids in starting material = 133.5 kg
% water in starting material = 94%
% solids in starting material = 100% – % water in starting material
= 100% -94%
= 6% solids
Let X kg be the weight of cherry tomatoes at the start 6% of the starting weight is solids, so 6% of X kg is equal to 133.5 kg of solids
0.06 X = 133.5 kg
X =133.5 kg/ 0.06
= 2,225 kg

9. The apple has a dry basis moisture content of 5.7 grams of water per gram of dry solids. What is its wet basis moisture content?
a) 85.0%
b) 88.0%
c) 99.8%
d) 60.8%

Explanation:
% Wet Basis Moisture = Weight of Water in the Sample * 100 / % of Total Weight of the Sample
Total weight of the sample = 5.7 g of water + 1.00 g of dry solids.
= 6.7 g
% Wet Basis Moisture = 8.9 * 100 / 9.9
= 85.0%

10. What weight of totally dry potato flakes would be obtained? If the potatoes weigh is 7676 grams after peeling, followed by drying. Its moisture content was 75% on a wet basis.
a) 1919 kg
b) 1999 kg
c) 2938 kg
d) 2002 kg

Explanation:
Weight of water = Weight of material * percentage water as a decimal
Weight of water in potato at starting of process = 7676 g * 0.75
= 5757 g
Weight of solids in potato at start = Weight of potato – weight of water
= 7676 -5757 g
= 1919 g

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