Question

base your answers to questions 7 through 10 on the information below and on your knowledge of earth science.
lake ontario’s seasonal density shift
lake ontario, a deep freshwater lake in new york state, experiences a crucial annual event driven by water’s unique properties. in the summer, warmer, less dense surface water floats on top of the colder, denser bottom water, creating a phenomenon called thermal stratification, where differences in temperature prevent water from mixing. in fall, as the surface cools, this water sinks, forcing the bottom water up. this causes the entire lake to turnover, or mix completely. this process is vital for replenishing dissolved oxygen in the deep waters, supporting aquatic life.
the following map shows isobaths, or lines connecting points of equal water depth in lake ontario.
water depth of lake ontario map

1. a scientist takes a 1,500 ml sample of water from the surface of lake ontario at the beginning of fall. the sample weighs 1,600 grams. calculate the density of this water, in g/ml, to the nearest hundredths and explain why this water sinks, leading to lake turnover.

density:
explanation:

1. calculate the water depth gradient from points c to d in lake ontario. show your work below and be sure to include units in your answer.

Explanation:

Question 7

Step1: Recall density formula

The formula for density is $
ho = \frac{m}{V}$, where $m$ is mass and $V$ is volume.

Step2: Substitute values

Given $m = 1600$ grams and $V = 1500$ mL. Substitute into the formula: $
ho=\frac{1600}{1500}\approx1.07$ g/mL.

Step3: Explain sinking

In fall, surface water cools. Cooler water is denser than the warmer, less dense bottom water (from summer stratification). When the surface water's density becomes greater than the underlying water's density, it sinks, causing turnover.

In fall, the surface water cools, increasing its density. When this water's density (1.07 g/mL) becomes greater than the density of the underlying water, it sinks, driving lake turnover.

Question 8

Step1: Recall gradient formula

Gradient is calculated as $\text{Gradient}=\frac{\text{Change in depth}}{\text{Change in distance}}$.

Step2: Determine depth change

From the map, point C is on the 100 m isobath, point D is on the 500 m isobath. So $\Delta\text{depth}=500 - 100 = 400$ m.

Step3: Determine distance change

Using the scale (10 miles = 16.0934 km = 16093.4 m, but here we use miles to miles first). The distance between C and D: from the scale, 1 unit is 10 miles. Visually, the distance between C and D is about 30 miles (estimating from the map's scale: 0 - 40 miles, and C to D is ~3 intervals of 10 miles). Convert miles to meters? Wait, no, let's use miles. Wait, the scale is 0 - 40 miles. Let's calculate the distance. The horizontal distance: using the scale, each segment is 10 miles. Let's say the distance between C and D is 30 miles (approximate from the map: C is at the top, D is in the middle; the scale bar: 0 to 40 miles. Let's measure the distance: from C to D, the length on the map: if the scale is 40 miles for the bar, and the distance between C and D is about 3 times 10 miles (since 10 miles is 1 segment). So distance $d = 30$ miles. Wait, but depth is in meters? Wait, no, the isobaths are in meters? Wait, the problem says "water depth", so isobaths are in meters? Wait, the map's isobaths: 100, 200, etc. So depth is in meters. The scale is in miles. So we need to convert miles to meters or keep units consistent. Wait, 1 mile = 1609.34 meters. But maybe the problem expects using miles and meters, but let's check:

Wait, the scale is 0 - 40 miles. Let's calculate the distance between C and D. Let's use the scale: the distance between C and D on the map: if the scale bar is 40 miles, and the distance between C and D is, say, 30 miles (visually, from the map, C is at the top, D is below; the horizontal distance: let's count the scale. Let's assume the distance between C and D is 30 miles (since 10 miles per segment, 3 segments). So $\Delta\text{distance}=30$ miles. Convert miles to meters: $30$ miles $\times1609.34$ m/mile $= 48280.2$ m. But wait, maybe the isobaths are in feet? No, the problem is Earth science, lake depth in meters or feet? Wait, Lake Ontario's depth: actual depth, but the map's isobaths: let's check the numbers: 100, 200, 300, 400, 500, 600, 700. So likely meters.

Wait, maybe the scale is in miles, and we can calculate the distance in miles, then convert depth change (in meters) to miles? No, better to use consistent units. Wait, maybe the depth is in feet? No, 700 feet is shallow for Lake Ontario. Wait, Lake Ontario's maximum depth is about 244 m, wait no, actual maximum depth is ~244 meters? Wait, no, Lake Ontario's average depth is 86 m, maximum 244 m. Wait, the map's isobaths go up to 700, which is inconsistent. Wait, maybe the isobaths are in feet. 700 feet is ~213 meters. Okay, let's proceed.

Wait, the problem says "water depth", so let's take the isobaths as depth in meters. The distance: using the scale, 10 miles = 16.0934 km = 16093.4 m. But let's use the scale in miles. Let's say the distance between C and D is 30 miles (from the map: the scale is 0 - 40 miles, and C to D is about 3/4 of the scale? Wait, no, looking at the map: point C is on the northern shore, point D is in the middle. The horizontal distance: using the scale, each 10 miles is a segment. Let's count the number of miles between C and D. Let's say the distance is 30 miles (as 3 segments of 10 miles). So $\Delta\text{distance}=30$ miles. Convert miles to meters: $30$ miles $\times1609.34$ m/mile $= 48280.2$ m. But d…

Answer:

Density: $\boldsymbol{1.07}$ g/mL