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Example 1.2: Using Displacement of Water to Determine Density

This PhET simulation illustrates another way to determine density, using displacement of water. Determine the density of the red and yellow blocks.

Solution

When you open the density simulation and select Same Mass, you can choose from several 5.00-kg colored blocks that you can drop into a tank containing 100.00 L water. The yellow block floats (it is less dense than water), and the water level rises to 105.00 L. While floating, the yellow block displaces 5.00 L water, an amount equal to the weight of the block. The red block sinks (it is more dense than water, which has density \( = 1.00\ \frac{\mathrm{kg}}{\mathrm{L}}\)), and the water level rises to 101.25 L.
The red block therefore displaces 1.25 L water, an amount equal to the volume of the block.

\(V_{\mathrm{red\ block}}\) \(= 1.25\ \mathrm{L}\)


\(m_{\mathrm{red\ block}}\) \(= 5.00\ \mathrm{kg}\)


The density of the red block is:

\(ρ_{\mathrm{red}}\) \(= \dfrac{m_{\mathrm{red\ block}}}{V_{\mathrm{red\ block}}}\)

\(\ \ \ =\dfrac{5.00\ \mathrm{kg}}{1.25\ \mathrm{L}}\)

\(\ \ \ =4.00\ \frac{\mathrm{kg}}{\mathrm{L}}\)


Note that since the yellow block is not completely submerged, you cannot determine its density from this information. But if you hold the yellow block on the bottom of the tank, the water level rises to 110.00 L, which means that it now displaces 10.00 L water, and its density can be found:

\(V_{\mathrm{yellow\ block}}\) \(= 10.00\ \mathrm{L}\)


\(m_{\mathrm{yellow\ block}}\) \(= 5.00\ \mathrm{kg}\)


\(ρ_{\mathrm{yellow}}\) \(= \dfrac{m_{\mathrm{yellow\ block}}}{V_{\mathrm{yellow\ block}}}\)

\(\ \ \ =\dfrac{5.00\ \mathrm{kg}}{10.00\ \mathrm{L}}\)

\(\ \ \ =0.500\ \frac{\mathrm{kg}}{\mathrm{L}}\)