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Using mass of copper in both air and liquid to find density of liquid?

  • #1
1. A piece of copper whose density is 8.93 gm/cm^3 weighs 180 gm in air and 162 gm when submerged in a certain liquid. What is the density of the liquid?



2. d = m/v



3. I wasn't sure how to compute this but I calculated the volume of the copper based on its density and air weight, then I just used that volume and the submerged weight to get 8.04 gm/cm^3 as the density of the liquid. Does this work and if so, can anyone clarify why? Thanks
 

Answers and Replies

  • #2
gneill
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I wasn't sure how to compute this but I calculated the volume of the copper based on its density and air weight, then I just used that volume and the submerged weight to get 8.04 gm/cm^3 as the density of the liquid. Does this work and if so, can anyone clarify why? Thanks[/b]
The density of air is about 1.2 x 10-3 gm/cm3. Compared to the copper's 8.04 gm/cm3 it will have a negligible effect on the outcome.

You might want to check your result. 8.04 gm/cm3 seems pretty high for your average liquid.
 
  • #3
I also found it to be rather high but I have not encountered such a problem, any tips as to how to understand this question is appreciated :)
 
  • #4
gneill
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Perhaps you could run through the calculations, step by step, that you performed to find the density of the liquid.
 
  • #5
I calculated the volume of the copper based on its density and air weight:

V = 180 gm/(8.93gm/cm^3) = 20.2 cm^3

Used that volume and the submerged weight:

d = 162gm/20.2 cm^3 = 8.02 gm/cm^3

I don't think I did it right because I'm not familiar with the physics of the situation but I'm trying to understand how to extract any understanding from the information given.
 
  • #6
gneill
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20,781
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I calculated the volume of the copper based on its density and air weight:

V = 180 gm/(8.93gm/cm^3) = 20.2 cm^3

Used that volume and the submerged weight:

d = 162gm/20.2 cm^3 = 8.02 gm/cm^3

I don't think I did it right because I'm not familiar with the physics of the situation but I'm trying to understand how to extract any understanding from the information given.
Okay. The purpose of the question is to show how one might use the effects of buoyancy to determine the density of an unknown liquid. The ideas involved include weight, density, volume displacement, and buoyant force.

You should review what material you have on buoyancy, but to cut to the chase as it were, when you immerse an object of known volume in a liquid, it displaces that volume of liquid (obviously). This displacement results in a buoyant force on the object, opposed to (in counter direction to) the force of gravity on the object. The magnitude of that force is equal to the weight of the liquid displaced by the object. If an object displaces its own weight in liquid, it floats (or in the borderline case where it just submerges when it has displaced its own weight in liquid, is said to be neutrally buoyant).

So, if you take an object with a known weight and volume and submerge it, its effective weight will be reduced by the force of buoyancy, which is equal to the weight of the volume of liquid displaced. The weight difference gives you the weight of the displaced volume of liquid. If you know the weight and volume of a sample of the liquid, you can determine its density.

When the object is relatively dense (compared to air) you can usually ignore the effect that buoyancy in air has on its measured weight.
 
  • #7
Wow that is very very interesting. Unfortunately this is one of my physics lab questions and the physics lab we are taking is completely separate from the physics course itself so the topics covered by our professor are not related strongly to the labs. For example I have no idea about the physics of buoyancy so that paragraph will help me immensely in solving this lab problem. :)
 
  • #8
I did the calculations and the only problem is that it seemed TOO simple haha. I got .90 gm/cm^3 which sounds a lot more reasonable! Thanks again!
 

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