Question

A tank in the shape of an inverted right circular cone has height 7 meters and radius 2 meters. It is filled with 5 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ=1100 kg/m3.

Answer 1

Given Information:

height of tank = 7 m

radius = 2 m

density = 1100 kg/m³

tank filled with hot chocolate = 5 m

Required Information:

work required to empty tank = ?

Answer:

work required to empty tank =  124789.93 J

Step-by-step explanation:

We can solve this problem using calculus.

The work required to empty the tank by pumping the hot chocolate is given by

W = F*d

Where F = mg

Consider an infinitesimal slice of cone with a height Δh, we can represent it in terms of the larger cone having a height of 7 m and radius 2 m

V = 1/3πr²Δh

where r = (2/7)*h

V = 1/3π((2/7)*h)²Δh

V = (4/147)πh²Δh

As we know

mass = density*volume

mass = 1100*(4/147)πh²Δh

mass = 29.932πh²Δh

F = mg

F = 9.8*29.932πh²Δh

F = 293.33πh²Δh

W = F*d

where d is (7 - h)

W = 293.33πh²Δh*(7 - h)

W = 293.33πΔh(7h² - h³)

Now we need to integrate the above expression with limits 5 to 0

W = ∫293.33π(7h² - h³)dh

W = 293.33π∫(7h² - h³)dh

W = 293.33π [7/3h³ - h⁴/4]

Evaluating limits

W = 293.33π [7/3(5)³ - (5)⁴/4]

W = 293.33π [291.667 - 156.25]

W = 124789.93 J

W = 124.78 kJ

Therefore, it would require 124.78 kJ of work to empty the tank by pumping the hot chocolate over the top of the tank.

Answer 2

Answer:

The work required to empty the tank is 1,580,887 J

Step-by-step explanation:

Work = mass of chocolate × acceleration due to gravity × height of chocolate

Volume of tank (cone) = 1/3πr^2h = 1/3 × 3.142 × 2^2 × 7 = 29.33 m^3

Mass of hot chocolate = density of hot chocolate × volume of cone = 1100 kg/m^3 × 29.33 m^3 = 32,263 kg

height of chocolate = 5 m

Work = 32,263×9.8×5 = 1,580,887 J