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
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.
