My goal is to monitor the change in specific gravity of a liquid over a period of time. My question is: What are the appropriate formula for determining expected apparent weight of an object immersed in a liquid where the liquids specific gravity g/ml is expected to change?

EG. If I were to take an object who's density is 2.6 (average for glass) weighing 100 grams and plunk it into distilled water I believe I should expect an apparent weight should be roughly 61.53 grams. Please let me know if I am just horridly wrong.

So then if that distilled waters density/specific gravity were to change say to 1.010, would my new apparent weight of the object be 61.15 grams?

My math is not solid in this. I'm basically using ratios in order to produce these answers. Please for the sake of simplicity if you are to choose to answer leave out extenuating circumstances such as temperature of the liquid/object and possible compression of the object due to pressure.

If you do chose to add extenuating circumstances I would ask to add those concepts as tertiary answers.

I'm sure that my question is probably very basic, but grasping the concepts has proven perplexing to me. I am probably not using the correct search. Your help in this simple question is greatly appreciated.

  • $\begingroup$ Note: specific gravity has no units; it is a dimensionless number. Density has dimensions of mass/volume. $\endgroup$ – DJohnM Mar 12 '14 at 7:09

Archimedes' principle tells us that the upwards force on an object immersed in a fluid is equal to the weight of fluid displaced.

So let's take your initial experiment. You don't tell us the volume of your object, but you do give its weight, $W_g = 100$ g, and density, $\rho_g = 2.6$ g/cm$^3$, so the volume is:

$$ V = \frac{W_g}{\rho_g} = 38.46 cm^3 $$

When you put this in water the volume of water displaced is the same as the above volume, so the mass of water displaced is:

$$ M_w = V\rho_w = \frac{W_g}{\rho_g} \rho_w = 38.46g $$

and hence the effective mass of your glass object is 100 - 38.46 = 61.54g as you say.

If the density of water changes to 1.010 g/cm$^{3}$ just use the equation above but set $\rho_w = 1.010$, and you will indeed get the effective weight of the glass equal to 61.15g. Though if you want to be really accurate you should note that changing the temperature will change the density of the glass as well, so $\rho_g$ would be slightly different as well.

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  • $\begingroup$ Thank you John. I appreciate your taking the time to help me out. $\endgroup$ – Nicknapoli82 Mar 12 '14 at 15:51

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