A1.0-m-diameter vat of liquid is 2.0 m deep. the pressure at the bottom of the vat is 1.3 atm. what is the mass of the liquid in the vat?

The mass of the liquid = 10538 kg

Explanation:

The pressure in a liquid is

P = ρgh Equation 1

ρ = P/gh Equation 2

Where P = pressure, ρ = density, g = acceleration due to gravity, h = height.

Given: p = 1.3 atm, h = 2.0 m, g = 9.81 m/s²

If,  1 atm = 1.013×10⁵ N/m²

Then, P = 1.3×1.013×10⁵ N/m² = 1.3169×10⁵ N/m²

Substituting in equation 2,

ρ  =  1.3169×10⁵/(9.81×2)

ρ =   1.3169×10⁵/19.62

 ρ = 6712.03 kg/m³.

But Density,

ρ  = m/v  

m =  ρ × v Equation 3

Where m = mass of the liquid, v = volume of the liquid in the vat

v = πd²h/4, where d = diameter = 1.0 m, h = 2.0 m.

v = 3.14(1)²×2/4

v = 1.57 m³ also,  ρ =  6712.03 kg/m³.

Substituting into equation 3

m = 1.57×6712.03

m = 10537.887 kg

m ≈ 10538 kg.

Thus the mass of the liquid = 10538 kg

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

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