# Answer to Question #34616 in Molecular Physics | Thermodynamics for Wendy

Question #34616

What is the mass of water required to fill a circular hot tub 2.4 meters in diameter and 1.2 meters deep answer in units of kilograms?

Expert's answer

m = ρ* V = ρ * S * H = ρ * Pi * (d^2) / 4 * H

Where:

ρ - density of water [ =

1000 kg / (m^3) ]

V - volume of circular hot tub

S - area of bases of circular hot tub [ = Pi * (d^2 / 4) ]

Pi ≈3.14

H = 1.2 m

d = 2.4 m

So,

m ≈ 1000 * 3,14 * (2.4^2) * 1.2 / 4 ≈ 5425,92

(kg)

ANSWER: m ≈ 5425,92 (kg)

Where:

ρ - density of water [ =

1000 kg / (m^3) ]

V - volume of circular hot tub

S - area of bases of circular hot tub [ = Pi * (d^2 / 4) ]

Pi ≈3.14

H = 1.2 m

d = 2.4 m

So,

m ≈ 1000 * 3,14 * (2.4^2) * 1.2 / 4 ≈ 5425,92

(kg)

ANSWER: m ≈ 5425,92 (kg)

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