How do I calculate the volume of an irregular container using styrofoam beads?

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Discussion Overview

The discussion centers around calculating the volume of an irregularly shaped container using the weight of styrofoam beads and water. Participants explore various methods and assumptions related to the problem, which involves practical application rather than theoretical derivation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the weights of a water-filled cylinder and a cylinder filled with beads, seeking to determine the volume of an unknown container based on these measurements.
  • Another participant suggests that the ratio of the weights of water to beads should remain consistent across containers, questioning whether this is a homework problem.
  • A participant expresses their lack of familiarity with the topic due to time away from academic studies, indicating a practical need for the solution in a work context.
  • One response proposes that if the weight of water is known, the volume can be calculated by dividing the weight by the density of water.
  • A later reply calculates an expected water volume based on the ratio of weights from the beads and water, suggesting a volume of approximately 94 liters, while noting the limitations of significant figures.
  • Another participant discusses the impact of container rigidity and temperature on the accuracy of the volume calculation, referencing specific density values for water at a given temperature.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions necessary for calculating the volume, particularly regarding the consistency of weight ratios and the impact of temperature on water density. The discussion remains unresolved with multiple competing approaches presented.

Contextual Notes

Assumptions about the consistency of weight ratios and the effects of temperature on water density are not universally accepted, and the calculations involve approximations that may introduce significant error.

tkettlehut
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I'm trying to find the volume of a irregularly shaped container by using very small styrofoam beads. Things I know: Cylinder A filled with water @ 18 c - net weight - 1231.0 g. Cylinder A filled with beads - net weight - 22.0 g. My unknown volume container holds 1678.7 g of beads. How the heck do I determine the volume of my container with this information?
 
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Of course, you are asked for an inexact solution. You need to make the assumption that the (weight of container filled with water) : (weight of container filled with Styrofoam beads) ratio is unchanged across containers. Are you sure this not a homework question?
 
I wish - I'm 23 years removed from college, (forgot more than I've learned) and I'm trying to get this answer for a problem at my job.
 
huh? i don't understand what information you have? you know how much the water weighs that fills the container?if you do then the volume of your container is that weight divided by the density of water.
 
tkettlehut said:
I'm trying to find the volume of a irregularly shaped container by using very small styrofoam beads. Things I know: Cylinder A filled with water @ 18 c - net weight - 1231.0 g. Cylinder A filled with beads - net weight - 22.0 g. My unknown volume container holds 1678.7 g of beads. How the heck do I determine the volume of my container with this information?

So you have

A: 22g of beads = 1231g of water
B: 1678.7g of beads

If the bags fill similarly, then we'd expect bag B to hold 1678.7g * (1231 / 22) = 93931g of water. Water is roughly 1 mL/g -- at 18 degrees I think it's a little more, but the error in the bead/water conversion is probably large enough that I wouldn't bother -- making the volume 93931 mL = 93.931 L. This probably has no more than 2 significant decimal places, so I'd report it as 94 liters of capacity.
 
If the containers are rigid (like a metal box instead of a backpack), then there's little enough error that dealing with the temperature makes sense. http://faculty.uccb.ns.ca/chowley/chem201/dh20vstemp.htm has densities for water based on temperature. In particular, it gives 0.998595 g/mL at 18 degrees C, giving volume per gram 1.00140698 mL/g. That gives a total volume of 94.063 liters.
 
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