Solve Simple Word Problem: 100kg w/15% Moisture to 5% Dry Weight Fibers

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Homework Help Overview

The discussion revolves around a word problem related to recycling paper, specifically calculating the amount of water needed to adjust the moisture content of a given mass of paper. The original poster presents a scenario involving 100 kg of paper with a moisture content of 15% and seeks to determine how much water must be added to achieve a solution that is 5% by dry weight fibers.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore various interpretations of the moisture content and the calculations involved in determining the amount of water to be added. Questions arise regarding the initial calculations of moisture and solid content, as well as the formulation of equations to represent the problem.

Discussion Status

There is an ongoing examination of the calculations and assumptions made by the original poster and other participants. Some participants provide alternative perspectives on how to set up the equations, while others express confusion about the relationships between the quantities involved. The discussion remains active with no clear consensus reached yet.

Contextual Notes

Participants note discrepancies in the initial calculations of moisture and solid mass, leading to further questioning of the setup and assumptions. The problem requires careful consideration of how the total mass and moisture percentages interact, and there is acknowledgment of the need for clarity in defining the variables used in the equations.

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Homework Statement


When paper is recycled, water is added to break the paper up into its original wood fibers. If 100 kg of paper with a moisture content of 15% is to be recycled, how much water (kg) must be added to the recycled paper to create a solution that is 5% by dry weight fibers?



Homework Equations


100kg at 15% moisture is 6.6kg water and 85% dry is 93.4 kg fibers
I want 95% moisture and 5% dry

The Attempt at a Solution


6.6kg+x=95kg
x=88.4 kg of water.
Is it that simple or have i missed something?
 
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Let's see. You have 93.4 kg fibers. To get a 5% solution, you would need X kg water. Since you already have 6.6 kg, you would have to add (X-6.6) kg water. Does that come out to be 88.4 kg?
 
How did you arrive at 15% of 100 kg is 6.6 kg and 85% of 100 kg is 93.4 kg?
 
Sorry. Obvious mistake. 100kg at 15% moisture give 15 kg moisture and 85 kg solid. I had originally divided 100kg by 15.
If I take my original solid mass of 85kg and add x amount of water, since I have 15kg of water already, its (x-15kg) I need to get a solution that is 5% or .05 water or 95% solid, so.
85kg+(x-15kg)=95kg
x=25

It seems I am missing something. If everything that goes in must equal everything that goes out I don't think I am accounting for the added water in my final mass of 95kg solid.
 
Last edited:
Okay, you already have 15 kg of water and you add x kg of water. No, you don't "need" x- 15. You now HAVE x+ 15 kg. of water. You still have 85 kg of solid so you know have a total of x+ 15+ 85= x+ 100 kg total. The percentage of that that is water is (x+ 15)/x+ 100= 0.05.

(If you wanted to add enough water to make a total of x kg of water, then you have to add x- 15 kg. But you said x is the amount of water you add (which is what the question asked), not the total amount of water.)
 
First, thank you. I am not trying to be difficult, I am really trying to understand this.
I tried the question again. I drew a diagram like this..


100kg
A(w).15==> (re- )==>.05(s) C
(s).85==> (cycler)==>.95(w)

^
^​
B


1.Now I have a water(w) balance of...
.15(100)+B=.95C

2.I have a solid(s) balance of...
.85(100)=.05C

From equation 2, I get C=1700.
Substituting 1700 into equation 1, I get B=1600

From the question, 1600Kg of water must be added to the recycled paper to create a solution that is 5% fibers.
Does that look reasonable or am I really out to lunch?
 

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