Question

In: Other

Asample of a food material weighing 10 kg is initially at 450% moisture content dry basis....

Asample of a food material weighing 10 kg is initially at 450% moisture content dry basis. It is dried to 10% moisture content wet basis. How much water is removed from the sample per kg of dry solids?

Solutions

Expert Solution

The important thing to learn in the question is ' Dry Basis ' and 'Wet Basis' calculation.So when equating the amount of moisture content in percentage,the numerator term remains always same which is the mass of water retained in the solid which is the food material in this case.But the denominator term varies for these 2 cases.For dry basis,we usually have mass of solid only in the numerator while for wet basis we have total mass of solid i.e. (mass of water + mass of dry solid). Wet basis is different from Dry basis in terms of mass addition of water to mass of solid for calculating amount of moisture.

Amanda K answered 5 months ago
Next > < Previous

Related Solutions

The following data have been collected for a dry food: Water Activity Moisture Content (g H2O...
The following data have been collected for a dry food: Water Activity Moisture Content (g H2O / g product) 0.1 0.060 0.2 0.085 0.3 0.110 0.4 0.122 0.5 0.125 0.6 0.148 0.7 0.173 0.8 0.232 (4 Point) This product was stored in an environment where after 30 days of storage the moisture content of the product (in dry basis), as measured in the laboratory, was 25%. If you were responsible for the warehouse and the quality of this product, would...
In batching of aggregates the moisture content is expected to be in saturated surface dry condition....
In batching of aggregates the moisture content is expected to be in saturated surface dry condition. However, in a concrete production, the actual moisture content of coarse aggregate on the site just before the batching measured as 8%, and its saturated surface dry condition requires 5% moisture content. If 1000kg/m3 of coarse aggregate and 220kg/m3 of free water is needed for a concrete production, how many kg of coarse aggregate and water is to be placed into the mixer for...
Following the soxhlet procedure, calculate the % fat content on dry and wet basis of a...
Following the soxhlet procedure, calculate the % fat content on dry and wet basis of a cheese sample considering the following: Weight of dried sample = 2.0251 g Weight of sample + thimble + glass wool = 5.4047 g Weight of defatted sample + thimble + glass wool = 4.4107 g % moisture = 35%
A compaction test measures a maximum dry density of 1666 kg/m3 and an optimum water content...
A compaction test measures a maximum dry density of 1666 kg/m3 and an optimum water content of 18.0%. The soil has an estimated Gs value of 2.68. A contract requires compaction to 95% of maximum dry density at water content of optimum or greater. A field test measures a moist density of 2023 kg/m3 and a water content of 27%. a) Use the values given for measured moist density and water content, calculate the field dry density b) Does the...
A 10 g bullet is fired with 450 m/s into a 10 kg block that sits...
A 10 g bullet is fired with 450 m/s into a 10 kg block that sits at rest on a wooden table 20 cm from the edge of the table. The bullet gets embedded in the block (perfectly inelastic collision). The block, with the embedded bullet, then slides to the edge of the table and drops down with some initial velocity while leaving the edge of the table. The coefficient of kinetic friction between the block and the surface of...
(a) An unstable particle with a mass equal to 3.34 ✕ 10−27 kg is initially at...
(a) An unstable particle with a mass equal to 3.34 ✕ 10−27 kg is initially at rest. The particle decays into two fragments that fly off with velocities of 0.977c and −0.851c, respectively. Find the masses of the fragments. (Hint: Conserve both mass–energy and momentum.) m(0.977c) = kg m(-0.851c) = kg (b) An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 4.00 ✕ 10−28 kg and that...
A countercurrent system is to treat (on an hourly basis) 10 tons of gangue (inert material),...
A countercurrent system is to treat (on an hourly basis) 10 tons of gangue (inert material), 1.2 tons of copper sulfate, and 0.5 tons of water with water as a fresh solvent. The solution produced is 10 percemt copper sulphate (remainder water). After each stage. 1 ton of inert retains two tons of water plus dissolved copper sulphate. How many stages are needed for a 98 percent of copper sulphate recovery? SEPA-LEACHING
An unstable nucleus with a mass of 16.3 × 10−27 kg initially at rest disintegrates into...
An unstable nucleus with a mass of 16.3 × 10−27 kg initially at rest disintegrates into three particles. One of the particles, of mass 4.9 × 10−27 kg, moves along the positive yaxis with a speed of 4.5 × 106 m/s. Another particle, of mass 8.7 × 10−27 kg, moves along the positive x-axis with a speed of 3.4 × 106 m/s. a) Find the speed of the third particle. Answer in units of m/s. b) At what angle does...
An unstable nucleus with a mass of 16.3 × 10−27 kg initially at rest disintegrates into...
An unstable nucleus with a mass of 16.3 × 10−27 kg initially at rest disintegrates into three particles. One of the particles, of mass 4.9 × 10−27 kg, moves along the positive yaxis with a speed of 4.5 × 106 m/s. Another particle, of mass 8.7 × 10−27 kg, moves along the positive x-axis with a speed of 3.4 × 106 m/s. a) Find the speed of the third particle. Answer in units of m/s. b) At what angle does...
An unstable particle with a mass equal to 3.34 ✕ 10−27 kg is initially at rest....
An unstable particle with a mass equal to 3.34 ✕ 10−27 kg is initially at rest. The particle decays into two fragments that fly off with velocities of 0.976c and −0.862c, respectively. Find the masses of the fragments. (Hint: Conserve both mass–energy and momentum.) m(0.976c) = kg m(-0.862c) =kg
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT