Question

A water truck's tank has internal dimensions of 11 meters in length and 1.5 meters in diameter. If the truck's load of pure water is 80 percent of full capacity, how heavy is the load in pounds? Water's density is 62.4 lb/ft3. In kilograms, what is the weight of the load if the truck carries liquid nitrogen instead of water? The liquid nitrogen's density is 0.808 g/ml.

Answer 1

Can you find the volume of the tank?

Answer 2

ive tried it a couple different ways and came out with the wrong answer every time

Answer 3

The tank is shaped like a cylinder. You need the formula for the volume of a cylinder.
The diameter of the base of the cylinder is 1.5 m, so the radius of the base is 0.75 m. The height of the cylinder is 11 m.

Answer 4

For a cylinder,

\(V = \pi r^2 h\)

where
V = volume
r = radius of the base
h = height

Answer 5

19.43860454=V correct?

Answer 6

then do you have to multiply by .80 for the 80 %

Answer 7

Correct.

Answer 8

so then you should get 15.55088364 right. Then how do you get the load in pounds?

Answer 9

Good. So far we know that the truck has 15.55 m^3 of water.
Since we are given the density of water in lb/ft^3, we need to find how many cubic feet we have.

Answer 10

The conversion factor from ft to m is:

\( 1~ft = 0.3048~ m\)

Answer 11

would you set it up
15.55m^3(1ft/0.3048)^3(62.4lb/ft^3)?

Answer 12

No. Don't do the density part now. Just do the conversion from cubic meters to cubic feet.
Be careful with one detail.
If 1 m = 0.3048 ft, then to find the conversion factor from cubic meter to cubic feet, you need to cube both sides:

Start with meters and feet.

\(1 ~m = 0.3048 ~ft\)

Now cube both sides:

\((1 ~m)^3 = (0.3048 ~ft)^3\)

Notice that 0.3048 must be cubed, so you get:

\(1~m^3 = 0.028317 ft^3\)

Answer 13

so then you do
15.55*0.028317*62.4?

Answer 14

A cubic meter is much larger than a cubic foot, so you divide 15.55 by the conversion factor 0.028317 to get cubic ft, since dividing by 0.028.. will give you a larger number. If you multiply 15.55 by 0.028... you'll get a smaller number. That can't be.

Answer 15

makes sense i got 34268.28898 for pounds per cubic foot, and i really appreciate the help, but what would the weight be if it were nitrogen?

Answer 16

Sorry. I'm redoing the conversion post above. I made a mistake above.

Just do the conversion from cubic meters to cubic feet.
Be careful with one detail.
If \(1~ ft = 0.3048 ~m\), then to find the conversion factor from cubic meter to cubic feet, you need to cube both sides: Start with meters and feet.

\(1 ft =0.3048 m\)

Now cube both sides:

\((1 ft)^3=(0.3048 m)^3\)

Notice that 0.3048 must be cubed, so you get: \(1~ ft^3=0.028317~m^3\)

Answer 17

Ok, now to find the cubic feet of water, we divide:

\(15.55 ~m^3 \div (0.028317~m^3/ft^3) \)

\( = 549.1 ~ft^3\)

Now we use the density.

549.1 ft^3 * 62.4 lb/ft^3 = 34,266 lb

That's more or less what you got.

Answer 18

Ok, for nitrogen, we need to deal with the density of nitrogen.
We already know the volume of the water in m^3 from before.
Now we have the same volume of nitrogen.

0.808 g/ml

1 ml = 1 cm^3

So the density of nitrogen is 0.808 g/cm^3

Also 1 m = 100 cm, so again cubing both sides:

1 m^3 = (100 cm)^3

1 m^3 = 1,000,000 cm^3

Answer 19

youd multiply them all together and divide by 1000 right?

Answer 20

i just tried it and it was right

Answer 21

thank you so much for all your help i really do appreciate it you really helped me understand the question and walked me through it. sorry if i was a bit slow its been a long day but thanks again.

Answer 22

We have 15.55 m^3 of nitrogen.
Multiply by 1,000,000 to find how many cm^3 we have
That is 15.55 * 1,000,000 = 15,550,000 cm^3
Now we use the density of nitrogen:

15,550,000 cm^3 * 0.808 g/cm^3 = 12,564,000 g

Now we convert to kg by dividing by 1000

12,564 kg