Work Required to Empty Tank of Hot Chocolate

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SUMMARY

The work required to empty a tank shaped like an inverted right circular cone, with a height of 8 meters and a radius of 6 meters filled with 7 meters of hot chocolate, is calculated using the formula W = ∫ (m * g) * (8 - h) dh. The mass of the hot chocolate is determined to be 54360π kg, and the density is 1510 kg/m³. The correct integral for calculating work is ∫ from 0 to 7 of (54360π * 1510 * (8 - h)) dh, resulting in a total work of approximately 8.229 x 10^6 Joules.

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A tank in the shape of an inverted right circular cone has height 8 meters and radius 6 meters. It is filled with 7 meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is 1510kg/m^3

m = pi*r^2
m = 36*pi*1510
m = 54360*pi
distance = (8-x)

[tex]\int_{0}^{7} 54360*pi*(8-x) dx[/tex]

this is clearly wrong, because i already tried to submit this, but it wouldn't accept it. can someone tell me what I am doing wrong?
 
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ProBasket said:
A tank in the shape of an inverted right circular cone has height 8 meters and radius 6 meters. It is filled with 7 meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is 1510kg/m^3
From problem statement:
{Height of Liquid Top Surface} = h
{Radius of Liquid Top Surface} = r = (6/8)*h = (3/4)*h
{Volume of Liquid in Tank} = (1/3)*π*r2*h = (1/3)*π{(3/4)*h}2*h = (0.589)*h3
{Mass of Liquid in Tank} = M = {Density}*{Volume of Liquid in Tank} =
= (1510)*(0.589)*h3 = (889.4)*h3

{dM/dh} = (3)*(889.4)*h2 = (2668)*h2
{Work Lifting Liquid Over Tank Top} = W = ∫ g*(8 - h)*(dM/dh)*dh

[tex]\ \ \ \ W = \int_{0}^{7} g(8 - h)(2668)h^{2} \ dh[/tex]

[tex]\ \ \ \ W = (2668)g\int_{0}^{7} (8 - h)h^{2} \ dh[/tex]

[tex]\ \ \ \ W = (26173)\int_{0}^{7} (8h^{2} - h^{3}) \ dh[/tex]

[tex]\ \ \ \ \color{red} W = (26173)(314.4) = (8229000 \ J) = (8.229*10^{6} \ J)[/tex]


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Your approach is correct, but there is a mistake in your integration. The correct integral should be:

\int_{0}^{7} 54360*pi*(8-x)*1510 dx

This is because the work required to pump out the hot chocolate is equal to the force (mass * gravity) multiplied by the distance it is pumped (8-x) and the density of the hot chocolate. So the formula should be:

Work = Force * Distance = (m*g) * (8-x) = (m*1510) * (8-x) = (36*pi*1510) * (8-x)

Then you integrate this formula from 0 to 7 (the initial and final heights of the hot chocolate) and you will get the correct answer.
 

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