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attashe74 [19]
3 years ago
11

Strawberries contain about 15 wt% solids and 85 wt% water. To make strawberry jam, crushed strawberries and sugar are mixed in a

45:55 mass ratio, and the mixture is heated to evaporate water until the residue contains one-third water by mass.
a) Draw and label a flowchart of this process.
b) Do the degree-of-freedom analysis and show that the system has zero degrees of freedom (i.e., the number of unknown process variables equals the number of equations relating them). If you have too many unknowns, think about what you might have forgotten to do.
c) Calculate how many pounds of strawberries are needed to make a pound of jam.
d) Making a pound of jam is something you could accomplish in your own kitchen (or maybe even a dorm room). However, a typical manufacturing line for jam might produce 1500 \mathrm{lb}_{\mathrm{m}} / \mathrm{h}. List technical and economic factors you would have to take into account as you scaled up this process from your kitchen to a commercial operation.
Chemistry
1 answer:
maxonik [38]3 years ago
8 0

Answer:

A. The flow diagram is in the attachment

B. Degree of freedom = 0

C. 0.486

D. Energy, power, cost of production

Explanation:

1/3 of water = 0.3333

Sugar = 1-1/3 = 2/3

B.) 45/55 = 9/11 when divided by 5

m1/m2 = 9/11

Cross multiply and make m1 subject of the formula

m1/m2 = 0.818

m1 = 0.818m2 ----1

0.15m1 + m2 = 1 x 0.667

0.15m1 + m2 = 0.667

0.15(0.818m2) = 0.667

1.1227m2 = 0.667

M2 = 0.5941

We put value of m2 in equation 1

M1 = 0.818 x 0.5941

M1 = 0.486

So 0.486 pound of strawberry is what we need in order to make 1 pound of jam.

D. We need great heat for water evaporation

The technical and economic factors would be

1. Energy which is power consumption

2. Cost of production for equipments and materials

3. Power

4. And hazards

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  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

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         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

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The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

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         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

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multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

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