What is the mass of the liquid in the vat?

Question

I have a 0.60-meter-diameter vat filled with liquid, and the vat is 2.7 meters deep. The pressure measured at the bottom of the vat is 1.5 atm. I attempted to solve this problem using the formula ( P = P_0 + \rho g d ), where I replaced the density (( \rho )) with mass divided by volume to solve for the mass. 
Here’s what I did:
- I used 151,987.5 Pa (equivalent to 1.5 atm) for ( P ).
- I assumed ( P_0 ) (the atmospheric pressure at the surface of the liquid) to be 101,325 Pa (equivalent to 1 atm). However, I’m starting to think this assumption might be incorrect.  
Despite my efforts, I got the wrong answer. I suspect my confusion comes from not knowing the correct values to use for the pressures in the formula. Could someone walk me through the correct steps to solve this problem? I'd greatly appreciate a detailed explanation. Thank you!

Anonymous · Accepted Answer

p=p0+&rho;ghthen&rho;=(p-p0)/(gh)1 atm=101,325 Pa&rho;=(1.5-1)*101,325/(9.8*2.7)&rho;=1914.7 kg/m^3m=&rho;VV=0.76 m^3m=1914.7*0.76=1455.2 kg

Anonymous · Answer

Get the volume and use the pressure to determine the density. 
What is missing is the pressure is absolute or relative. I'll assume absolute.
V = πr²h  = π(0.3)²(2.7) = 0.763 m³
Pfluid = ρgh 
Pfluid is pressure in Pa or N/m² 
ρ is the density of the fluid in kg/m³ 
g is the acceleration of gravity 9.8 m/s² 
h is the height of the fluid above the object in m 
1.5 atm is 1.5•101 kPa
ρ  = P/gh = (151.5e3) / (9.8)(2.7) = 5726 kg/m³
(possibly molten tin)
5726 kg/m³ x 0.763 m³ = 4370 kg

Anonymous · Answer

the pressure times the area of the bottom of the vat = the total force of gravity on the mass
this is a snap to calculate

Anonymous

What is the mass of the liquid in the vat?

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