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lauric acid is the main fatty acid in coconut oil and is currently being studied for potential health benefits. this white, powdery substance remains solid at room temperature (23°c). to investigate the physical properties of lauric acid, students added 15 g of lauric acid and 15 g of water to a test tube and placed it in a 60°c water bath (see figure 1) to melt the sample.
figure 1
part a
when the test tube registered 60°c, the students removed the test tube from the heated beaker and placed it in a second water bath that was 23°c (see figure 1). they recorded the temperature of the sample every 20 seconds until the sample solidified. results are shown in figure 2.
figure 2
part b
when the lauric acid sample reached 38°c, the students removed it from the 30°c water bath and placed it back in the 60°c bath. they recorded the temperature every 20 seconds for 400 seconds. the results are shown in figure 3.
figure 3
which of the following conclusions is best supported by the experiment?
title
lauric acid
your result
correct
difficulty
very hard
your pace
0:03
others' pace
1:20
text explanation
the question gives us multiple conclusions and asks us to identify the option that is consistent with the experiment. option (a) states that it takes less time for lauric acid to cool from 50-45 than to warm from 45-50 and option (b) states the opposite. without even looking at the graphs, we need to eliminate options (a) and (b) because our experiment only shows us the warming and cooling times of lauric acid when the water baths were 60 degrees and 23 degrees. if the baths had been extremely hot or extremely cold, lauric acid would have warmed or cooled more rapidly.
option (c) is incorrect because lauric acid cools more slowly from 55 to 50 degrees than from 50 to 45 degrees.
the correct answer is (d): under constant temperature conditions, lauric acid cools more quickly between temperatures 50 and 45 degrees than between 45 and 40 degrees. even though we only know the cooling times when the water bath is at the constant temperature of 23 degrees, a colder water bath would only speed up the overall cooling process, it wouldn’t change the cooling pattern. we would still see temperature drops that were steeper from 50 to 45 degrees than from 45-40. this makes sense when we remember that the freezing point of lauric acid is 44 degrees and the change from liquid to solid might slow down the speed of its cooling.
related lessons
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