How long will it take to fill a cube-shaped vat with edge length 34.7 ft with a certain liquid (d=0.504 g/mL) if it is spilling out of the hose at 12641.7 g/s? (1.00 in = 2.54E0 cm)

Question

A. 1.63×10^-8 hr

B. 1.17×10⁸ hr

C. 7.79×10⁶ hr

D. 0.212 hr

E. 13.1 hr

Answer 1

We will approach the problem by separating the question into three separate parts. First, we need to find the total volume of the container. Next, we need to calculate the total mass of the liquid needed to fill the volume of the container. Last, we will calculate the amount of time needed for the hose to fill the container using the total liquid's mass needed.

Step 1: What is the volume of the container we are going to fill?

The container is a cube. To find the volume of a cube we will use the equation,

V= (edge length)³

edge length of the cube = 34.7 ft

To prepare us for future calculations for this problem, the best ending units to be in for step one will be cubic centimeters to prepare us to use the density of the liquid in step 2 that has derived units that includes milliliters. Knowing 1 mL = 1 cm³ will be helpful as well.

Using the conversion factors, 1 ft = 12 in and 1 in = 2.54 cm, we will convert 34.7 ft to cm and then use the equation to find the volume of a cube to have ending units of cubic centimeters to use in step 2.

(34.7 ft/1) x (12 in/1 ft) x (2.54 cm/ 1 in) = 1,058 cm

V = (1,058 cm)³ = 1.18 x 10⁹ cm³

1 cm³ = 1 mL so 1.18 x 10⁹ cm³ = 1.18 x 10⁹ mL

Step 2: What is the mass of the liquid needed to fill the container's volume?

Using the volume of the container found in step 1, in units of cubic centimeters, we can find the total mass of the liquid needed to fill the container's volume using the liquid's given density of 0.504 g/ 1 mL.

(1.18 x 10⁹ mL/1) x (0.504 g/ 1 mL) = 5.96 x 10⁸ grams of liquid needed to fill the container's total volume

Step 3: What is the time needed for the liquid in the hose to transfer the mass needed to fill the container?

Using the total mass of the liquid found in step 2 that will fill the container's volume, we can find the time required for the hose to transfer the liquid's total mass needed to fill the volume of the container using the information given that the hose functions at a rate of 12,641.7 g/ 1 s.

(5.96 x 10⁸ grams of liquid needed to fill the container's total volume/1) x (1 s/ 12,641.7 grams of liquid) = 4.72 x 10⁴ seconds for the hose to transfer the required mass of the liquid to fill the container's total volume

(4.72 x 10⁴ seconds/1) x (1 hr/ 3600 s) = 13.1 hrs