Variables:

Controlled: 1.Type of matches2.Amount of air

Manipulated: 1.Temperature of air

Responding: 1.How long a match burns2.How much a match burns

Procedure

1. Find a plate and place a “golf ball size” amount of clay on the plate with some water.

2. Find a quart jar, box of matches, timer, ruler, pencil, and paper.

3. Calculate the volume of the jar and record.4. Light a match and time how long it takes to burn, repeat

15 times, and average the results.5. Record the average from Step 4 as Control Average.6. Light a match and place it in the clay on the plate, push

jar into clay around the match, and time how long it takes for the match to extinguish itself.

7. Remove the match intact and measure the length of the burnt part of the match using the ruler.

8. Record the results from Steps 6 and 7 as Inside Data.9. Repeat Steps 6, 7, and 8 15 times; average the time and

length results and record as Inside Data.10.Repeat Steps 6, 7, 8, and 9 outside and replace the name

record averages as Outside Data.11.Calculate how long it would take for a match to burn

inside and outside in the same circumstances with the data you have collected.

12.Should any of your times approach the Control Average, redo the test.

Conclusions If you burn a match in warm air, and burn a match in cold air, they’ll both consume the same amount of air, and burn at the same rate. However, if you burn two separate matches in two closed systems with limited air, A and B. Were A has warm air, and B has cold air, the B match will burn longer. This is because B’s air is denser, meaning there's more air per unit of volume. In my investigation the average time the match remained lit in cold air was 9.6 seconds, while warm air was 7.3 seconds. Without a closed system a match burning in warm air will just consume a greater volume of air than a match in cold air. My hypothesis was incorrect when I said “I predict that the match will burn less air,” but correct when I said “it will burn longer in the colder air”. Knowing that air is denser when it’s colder helps me to better understand how air flows through our valley and other natural phenomena.

ResultsControl:

Inside:

Outside:

Abstract“How much air will a match consume at different temperatures?” That was the question I was trying to answer in my investigation. To test this, I let a match burn in a closed system indoors and outdoors (during the winter) until it extinguished itself. I timed how long it took the match to burn until extinguished, and after each test, I measured the percentage of the match used. My hypothesis was that the match would burn less air and for a longer period of time in colder air because the air is denser. However, after analyzing my collected data, I disagree with my hypothesis. The results of this experiment indicate that in a closed system where air is limited, a match will burn longer in cold air because there is more air per unit of volume.

HOW MUCH AIR WILL A MATCH CONSUME AT DIFFERENT TEMPERATURES?

MATERIALS USED:

•Matches•Timer•Clay•Quart Jar•Plate•Water•Thermometer•Ruler•Pencil•Paper•Camera

Figure 1. Average percentage of match burned and average time to extinguish for experiments conducted inside (gray) and outside (yellow).

Validity•I repeated each test 15 times.•I used small increments on the ruler to be as precise as possible.•I made sure not to breath too hard on the matches.

One change I could make in my investigation would be to see what would happen if I ran a series of tests in the middle of summer and in our chest freezer. By changing the temperature by so much, I think I would see that the difference between “average time to burn” would greatly increase.

HypothesisI PREDICT THAT THE MATCH WILL BURN LESS AIR AND LONGER IN THE COLDER AIR; BECAUSE THE COLDER AIR IS DENSER, MEANING THEIRS MORE AIR PER UNIT OF VOLUME.

Step 1:

Step 7:Step 6:

Step 4:

My Smokey, Gaseous Investigation By: Logan Davis, Morgan Middle School, 8th grade Properties of Matter 05/17/2012

Test # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Time (sec.)

61 35 8 31 44 54 40 57 37 19 65 65 14 70 43

Average Time to Burn: 42.9 sec.

Test # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Time (sec.)

12 9 13 4 10 11 2 4 2 6 7 10 7 9 3

Length (in.)

.44 .44 .75 .19 .5 .63 .13 .25 .38 .38 .31 .44 .38 .44 .19

Average Time to Burn: 7.3 sec. Average Length Burned: 0.39 in.

Test # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Time (sec.)

10 12 10 12 13 12 12 10 5 5 11 6 9 12 6

Length (in.)

.75 .63 .38 .56 .81 .56 .5 .5 .13 .31 .5 .13 .25 .56 .19

Average Time to Burn: 9.6 sec. Average Length Burned: 0.45 in.

0

5

10

15

20

25

30

35

40

45

Amount Burned (in %) Time to Burn (in sec.)

Inside

Outside

•Outside below freezing, a match burned 143.5 cm3 of air in an average of 9.6 sec, and used 45% of itself. This is how I found how much air one match would burn:100 � 45 = 2.2m 2.2m � 143.5a = 315.7TA 2.2m � 9.6t = 21.12TT

So one match to burn fully would require (on average) 315.7 cm3 of air, and it would burn in 21.12 sec.

•Inside at room temperature, a match burned 143.5 cm3 of air in an average of 7.3 sec, and used 39% of itself. This is how I found how much air one match would burn:•100 � 39 = 2.6m 2.6m � 143.5a = 373.1TA 2.6m � 7.3t = 18.98TT

So one match to burn fully would require (on average) 373.1 cm3 of air, and it would burn in 18.98 sec.