Problems #11-25

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**Problem #11:** A material will float on the surface of a liquid if the material has a density less than that of a liquid. Given that the density of water is approximately 1.0 g/mL, will a block of material having a volume of
1.2 x 10^{4} in^{3} and weighing 350 lb float or sink when replaced in a reservoir of water?

**Solution:**

1) Convert cubic inches to cubic centimeters, then to mL:

1 in^{3}= 1 in x 1 in x 1 insince 1 inch = 2.54 cm:

1 in^{3}= 2.54 cm x 2.54 cm x 2.54 cm = 16.387 cm^{3}since 1 mL = 1 cm

^{3}, we know this:16.387 cm^{3}= 16.387 mL

2) Convert the cubic inches of the block of material to mL:

1.2 x 10^{4}in^{3}x 16.387 mL/in^{3}= 196644 mLAn alternate formulation of this conversion is:

1.2 x 10

^{4}in^{3}x (2.54 cm/in)^{3}= 196644 cm^{3}and then convert cm

^{3}to mL.

3) Convert pounds to grams:

350 lb x 453.6 g lb¯^{1}= 158760 g

4) Calculate the density of the material and compare it to the density of water:

158760 g / 196644 ml = 0.807 g/mLSince this is less than the density of water, we conclude that the block of material will float.

**Problem #12:** Gasoline has a density of 0.76 g/mL. Find the mass of 4.25 gallons in pounds.

**Solution:**

1) Convert g/mL to pounds per mL:

0.76 g/mL x (one lb / 453.59 g) = 0.0016755 lb/mL

Convert lb/mL to lb/gal:

0.0016755 lb/mL x (3 785.4 mL/1 gal) = 6.34244 lb/galTry this in Google: convert 0.76 g/mL to pounds/gallon

3) Find the mass of 4.25 gal:

6.34244 lb/gal x 4.25 gal = 27.0 pounds

**Problem #13:** The density of water is 1.00 g/cm^{3}. What is the density of water expressed in lb/ft^{3}?

**Solution:**

1) Convert 1.00 ft^{3} to cm^{3}:

one foot = 12 inches (by definition)12 inches times (2.54 cm/inch) = 30.48 cm

30.48 cm x 30.48 cm x 30.48 cm = 28316.8466 cm

^{3}

2) Convert 28316.8466 cm^{3} to grams, then pounds:

28316.8466 cm^{3}x 1.00 g/cm^{3}= 28316.8466 g28316.8466 grams divided by 453.59237 grams/pound = 62.43 pounds

There are 62.4 lb/ft

^{3}(to three sig figs)

**Problem #14:** Suppose a room is 11.8 ft wide by 12 ft long and 8.5 ft high and has an air conditioner that exhanges air at a rate of 1200 L/min. How long would it take the air conditioner to exhange the air in the room once?

**Solution:**

1) Convert feet to decimeters:

from Google, one foot = 3.048 dmtherefore:

11.8 ft times 3.048 dm/ft = 35. 9664 dm

12 ft times 3.048 dm/ft = 36.576 dm

8.5 ft times 3.048 dm/ft = 25.908 dm

2) Calculate the volume in cubic decimeters:

35. 9664 dm times 36.576 dm times 25.908 dm = 34082.156 dm^{3}

3) Compute time required:

34082.156 dm^{3}/ 1200 L/min = 28.40 minRemember, L = dm

^{3}, so those two units cancel exactly.

**Problem #15:** Convert 22.43 gal/min to L/seconds.

**Solution:**

1) Convert gallons to liters:

22.43 gal/min x (3.7854 L/gal) = 84.9067863 L/minAll I did when I typed up this solution was go to Google and type "convert 22.43 gal to L" and it gave me the answer I pasted in above.

2) Convert minutes to seconds:

84.9067863 L/min x (1 min / 60 sec) = 1.415 L/sec

**Problem #16:** Convert 35.50 mi/hr to km/s.

**Solution:**

1) Convert hours to seconds:

35.5 mi/hr times (1 hr / 3600 s) = 0.009861 mi/s

2) Convert miles to km:

0.009861 mi/s times (1.609 km / mi) = 0.01587 km/s

3) Click this sentence. It will go to a Google conversion of this problem.

**Problem #17:** Analysis of an air sample reveals that it contains 3.5 x 10^{-6} g/L of carbon monoxide. Express the concentration of carbon monoxide in lb/ft^{3}. (1 lb = 453.6 g, 1 m = 3.27 ft)

**Solution:**

1) Convert grams to pounds:

3.5 x 10^{-6}g/L times (1 lb/453.6 g) = 7.716 x10^{-9}lb/L

2) Construct the L to ft^{3} conversion:

we know this by definition: 1 L = 1 dm^{3}we know this: 1 dm

^{3}= 1 dm x 1 dm x 1 dmexpressed using meters: 1 dm

^{3}= 0.1 m x 0.1 m x 0.1 mwe know this: 0.1 m = 0.327 ft

Therefore: 1 L = 0.327 ft x 0.327 ft x 0.327 ft = 0.03496 ft

^{3}

3) Convert L to cubic feet:

7.716 x10^{-9}lb/L times (1 L / 0.03496 ft^{3}) = 2.2 x 10^{-7}lb/ft^{3}(to two sig figs)

**Problem #18:** The formula for the volume of a sphere is given by: volume = (4/3)(3.14)r^{3}, where r is the radius. Calculate the weight in tons of a sphere of water having a radius of 1.00 mile. The density of water is 1.00 g/cm^{3}

**Solution:**

1) This converts mile to cm:

(1.00 mile) (5280 feet / mile) (12 inch / ft) (2.54 cm / inch) = 160934.4 cmNote: no rounding off until the final answer

2) This is the volume calculation:

V = (4/3) (3.14) (160934.4 cm)^{3}= 1.74508 x 10^{16}cm^{3}

3) Convert the volume to weight:

(1.74508 x 10^{16}cm^{3}) (1.00 g / cm^{3}) (1 lb / 453.6 g) (1 ton / 2000 lb) = 1.92 x 10^{10}tons (rounded to three sig figs)

**Problem #19:** Convert 66 ft/s to miles/hr.

**Solution:**

1) Convert feet to miles:

66 ft/s times (1 mi / 5280 ft) = 0.0125 mi/s

Convert seconds to hours:

0.0125 mi/s times (3600 s / hr) = 45 mi/hr

**Problem #20:** 0.211 kcal/yd^{2} in cal/in^{2}

**Solution:**

1) 1 kcal = 1000 cal. The calorie is a unit of heat.

0.211 kcal/yd^{2}times (1000 cal / kcal) = 211 cal/yd^{2}

2) Create the yd^{2} to in^{2} conversion factor:

1 yd^{2}= 1 yd x 1 yd1 yd = 3 feet and 12 inches in a foot, so 36 inches in a yard

1 yd

^{2}= 36 in x 36 in = 1296 in^{2}

3) Use it:

211 cal/yd^{2}times ( 1 yd^{2}/ 1296 in^{2}) = 0.163 cal/in^{2}(to three sig figs)

**Problem #21:** Convert 1.61 yd^{3} into cubic decameters

**Solution:**

1) The prefix for deca- is da- and it multiplies the base unit by 10. It is not a commonly-used unit in chemistry:

1 decameter (dam) = 10 metersSince 1 m = 1.09361 yd, we have this:

1 dam = 10.9361 yd

2) We need to cube the conversion factor:

1.61 yd^{3}x (1 dam / 10.9361 yd)^{3}= 0.00123 dam^{3}

**Problem #22:** The men's world record for running a mile outdoor is 3:44.39. At this rate, how long it will take a man to run a 1500 meter race?

**Solution:**

1 mile = 1609.34 m3:44.39 sec = 224.39 sec (from 3 min x 60 s/min plus 44.39 s)

224.39 sec is to 1609.34 m as x is to 1500

x = 209.14 sec

We know that 3 minutes is 180 seconds.

209.14 - 180 = 29.14 s

3:29.14

Compare to the current 1500 m record of 3:26.00. By the way, both records are held by the same person, Hicham El Guerrouj (of Morocco).

**Problem #23:** Convert 10,000 meters to miles.

1) Convert meters to centimeters:

10000 m x (100 cm/m) = 1,000,000 cm

2) Convert cm to miles via dimensional analysis. I will build up the conversion one step at a time:

1 x 10^{6}cm x (1 inch / 2.54 cm) <--- convert cm to inch1 x 10

^{6}cm x (1 inch / 2.54 cm) x (1 foot / 12 inch) <--- convert inch to foot1 x 10

^{6}cm x (1 inch / 2.54 cm) x 1 foor / 12 inch) x (1 mile / 5280 feet) <--- convert feet to milesThe answer is 6.214 miles (pay attention to sig figs if you face this problem in your class)

As of August 2014 (when this question was formatted), Kenenisa Bekele of Ethiopia is the world-record holder in the 10,000 meter run. You might want to look him up. He is the winner of four Olympic medals and six World Championship medals. He also holds the world record for the 5,000 meter run. One of the better athletes this world has ever produced.

**Problem #24:** The density of mercury is 13.534 g/cm^{3}. What is the density in lbs/in^{3}?

**Solution:**

1) Convert grams to pounds:

1 pound = 453.613.534 g/cm

^{3}times (1 lb / 453.6 g) = 0.0298369 lb / cm^{3}

2) Convert cm^{3} to inch^{3}:

1 inch = 2.54 cm (by definition)1 in

^{3}= 1 in x 1 in x 1 in = 2.54 cm x 2.54 cm x 2.54 cm1 in

^{3}= 16.387 cm^{3}

3) Do the final step:

(0.0298369 lb / cm^{3}) times (16.387 cm^{3}/ 1 in^{3}) = 0.489 lb / in^{3}Comment: using Google calculator, you can always check your answer.

**Problem #25:** If the density of a substance is 1.43 lb/ft^{3}, find the mass in grams of 4.35 in^{3} of the substance.

**Solution:**

1) Let's change lb/ft^{3} to g/in^{3}. First the lb:

1.43 lb/ft^{3}times 453.59 g/lb = 648.6337 g/ft^{3}

2) Next is the change from ft^{3} to in^{3}. Start by thinking of ft^{3} this way:

1 ft^{3}= 1 ft by 1 ft by 1 ftsince 1 ft = 12 in, we have this:

1 ft

^{3}= 12 in by 12 in by 12 in = 1728 in^{3}Then, this:

648.6337 g/ft

^{3}times 1 ft^{3}/ 1728 in^{3}= 0.375367 g/in^{3}Often, the conversion just above is shown this way:

648.6337 g/ft

^{3}times (1 ft / 12 inches)^{3}= 0.375367 g/in^{3}

3) Last step:

0.375367 g/in^{3}times 4.35 in^{3}= 1.632845 gto three sig figs, 1.63 g

**Bonus Problem:**In setting the world record for the 10,000 meter run, Kenenisa Bekele of Ethiopia averaged 14.18 miles per hour. Calculate his approximate time (minutes, seconds, tenth and hundredth of a second).

**Solution:**

There are a variety of ways to attack this problem. Mine is not the only way to an answer.

1) I'm going to convert 10,000 meters into miles, but I'm not going to use one single factor, like 1 m = 0.000621371 mile. I'm going to do it in several steps.

39.3701 inches 1 foot 1 mile 10,000 m x –––––––––––– x ––––––– x ––––––– = 6.2137 mile 1 m 12 inches 5280 feet

2) I'm going to add one more factor to the above:

39.3701 inches 1 foot 1 mile 1 hr 10,000 m x –––––––––––– x ––––––– x ––––––– x ––––––––– = 0.43820276994 hr 1 m 12 inches 5280 feet 14.18 miles

3) What I will do now is convert hours to seconds:

(0.43820276994 hr) (3600 s / hr) = 1577.52997 s

4) Now, I have to split out minutes from the 1577.53 seconds:

1577.53 s / 60 s/min = 26.292167 min

5) Determine how many seconds are in 0.292167 min:

(0.292167 min) (60 s / min) = 17.53 s

6) The calculation shows Bekele's time to be 26 min 17.53 sec. This, as it turns out, is his actual time from the 2005 race.

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