Question

How long will it take to fill a cylinder-shaped vat (with radius of 16.3 ft and depth of 19.0 ft) with a certain liquid (d=2.21 g/mL) if it is spilling out of the hose at 11587.4 g/s?
(1.00 in = 2.54E0 cm)

A.) 7.62×10⁸ hr

B.) 1.23×10⁷ hr

C.) 23.8 hr

D.) 0.385 hr

E.) 2.97×10^-8 hr

Answer 1

Answer 2

To calculate the time required to fill the cylinder-shaped vat:

Step 1: Convert dimensions from feet to centimeters

• Radius:

$$16.3 \, \text{ft} \times 30.48 \, \text{cm/ft} = 496.224 \, \text{cm}$$

• Depth (height):

$$19.0 \, \text{ft} \times 30.48 \, \text{cm/ft} = 579.12 \, \text{cm}$$

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Step 2: Calculate the volume of the cylinder

The formula for the volume of a cylinder is:

$$V = \pi r^2 h$$$$V = 3.1416 \times (496.224)^2 \times 579.12$$$$V \approx 3.1416 \times 246238.34 \times 579.12 \approx 4.474 \times 10^8 \, \text{cm}^3$$

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Step 3: Convert volume to mass

The liquid has a density of:

$$d = 2.21 \, \text{g/mL} \quad \text{(since 1 mL = 1 cm}^3\text{)}$$$$\text{Mass of liquid} = V \times d = 4.474 \times 10^8 \times 2.21 \approx 9.894 \times 10^8 \, \text{g}$$

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Step 4: Calculate time to fill

The liquid is spilling at a rate of:

$$11587.4 \, \text{g/s}$$$$\text{Time required: } t = \frac{\text{Mass}}{\text{Flow rate}} = \frac{9.894 \times 10^8}{11587.4} \approx 8.54 \times 10^4 \, \text{seconds}$$

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Step 5: Convert seconds to hours

$$t = \frac{8.54 \times 10^4}{3600} \approx 23.8 \, \text{hours}$$

✅ Correct answer:
C.) 23.8 hr.