Understanding Specific Gravity

Imagine you’ve got a handful of different objects and a bucket of water. You drop each object into the bucket, and some float while others sink. This whole sink-or-float business is essentially specific gravity at play.

Specific gravity is a handy trick for making comparisons in the material world. It’s a way of asking, “How does this material stack up against water?” The big formula here is a simple one:

\[ \text{SG} = \frac{\rho_{\text{substance}}}{\rho_{\text{water}}} \]

Where \( \text{SG} \) is the specific gravity, \( \rho_{\text{substance}} \) is the density of the substance, and \( \rho_{\text{water}} \) is the density of water.

Now, since water has a density of 1 gram per cubic centimeter (g/cm³), or 1000 kilograms per cubic meter (kg/m³), the math becomes pretty friendly. If the SG of a substance is 0.8, that means it’s 80% as dense as water.

So, if you know the specific gravity of an object, finding its density is straightforward. Just multiply the specific gravity by the density of water. For instance:

\[ \rho_{\text{substance}} = \text{SG} \times \rho_{\text{water}} \]

If an object has a specific gravity of 2:

\[ \rho_{\text{substance}} = 2 \times 1000 \, \text{kg/m}^3 = 2000 \, \text{kg/m}^3 \]

And what about the floating and sinking part? If the SG is less than 1, the object floats, and the percentage that’s submerged equals the SG. So, at an SG of 0.8, 80% of the object is underwater, and 20% peeks out like an iceberg. If the SG is more than 1, down it goes to sleep with the fishes.

That’s the gist of specific gravity! It’s a handy way to make a quick judgment about whether an object will float or sink in water, and with a little multiplication, it can tell you the density of the material. Let’s look at how the MCAT tends to ask questions on this concept and how to solve them.

How specific gravity comes up on the MCAT

Specific gravity might not show up all the time on the MCAT, but it’s still important to know. Typically, it’s featured in two main types of questions. The first is when you need to determine the density of an object because it’s part of another calculation. The second type of question focuses on how much of an object is submerged or above the water, which could be asked in terms of percentage or volume. Let’s go through a systematic approach for each scenario.

Finding the Density of an Object

1. Understand the Question: Ensure the question asks for the density of an object using its specific gravity.
2. Recall Specific Gravity Formula: Specific Gravity (SG) = \(\frac{\text{Density of the substance}}{\text{Density of water}}\).
3. Identify Given Values: Determine the specific gravity provided and remember that the density of water is 1 g/cm³ or 1000 kg/m³.
4. Calculate Density: Multiply the specific gravity by the density of water.
   – Density of the substance = SG \(\times\) Density of water
   – This means if the density of water is 1 g/cm³ you don’t need to calculate anything because SG = density of the substance

Above Below Questions

1. Understand the Variation: Identify what the question asks for—percentage below/above water, or volume below/above water.
2. Use Specific Gravity for Percentage: Remember that the specific gravity of an object is the same as the percentage of the object that will be submerged in water (in decimal form).
   – If SG < 1: The object float, the percentage submerged = SG \(\times\) 100.
   – If SG > 1: The object sinks, it will be 100% submerged
3. Calculate Percentages:
   – For percentage below water: Percentage submerged = SG \(\times\) 100.
   – For percentage above water: Subtract the percentage below from 100.
4. Volume Calculations:
   – For volume below water: Volume submerged = Total volume of the object \(\times\) SG.
   – For volume above water: Volume above water = Total volume of the object \(\times\) percentage below.

Example Above Below Question

Navigating the different variations of questions regarding objects in water might feel a bit overwhelming at first. To get a handle on it, let’s work through an example that covers several of these twists, solving for multiple values to demystify the process.

Question: A wooden block with a specific gravity of 0.5 is floating in water. What percentage of the block is submerged? What is the volume of the wood above the water if the total volume of the block is 800 cm³?

Answer:

1. Percentage Submerged: \( SG \times 100 = 0.5 \times 100 = 50\% \)
2. Volume Submerged: \( 800 \, \text{cm}^3 \times SG = 800 \, \text{cm}^3 \times 0.5 = 400 \, \text{cm}^3 \) this also make sense because that is half of the original volume and we knew that 50% of the cube was underwater
3. Volume Above Water: If we know the volume underwater then it is super easy to find the volume above water since it will be equal to the total volume minus the volume underwater or 400 cm³ because \( 800 \, \text{cm}^3 – 400 \, \text{cm}^3 = 400 \, \text{cm}^3 \)

Thus, 50% of the block is submerged, and 400 cm³ of the wood is above the water.

Always read MCAT questions carefully to understand what is being asked, especially for the second type of question, as the problem could be phrased in various ways to test your understanding of specific gravity and related concepts.

Now that we’ve walked through how to approach specific gravity questions, it’s time to put your knowledge into practice. Give it a go!