Question

A large storage tank holds 2900 m3 of a liquid mixture that has 112% of the density of water. The tank is in an empty warehouse with a 40,000 ft2 of empty floor space. If over the weekend the tank leaks and empties onto the floor: a.) How deep is the liquid on Monday morning? b.) The liquid gets pumped out and carried away by trucks that can carry 9,000 kg of liquid each. How many truck loads are required?

Answer 1

Answer:

Part a) The height that the liquid forms equals 2.56 feet.

Part b) The number of trucks required are 361.

Step-by-step explanation:

When all the liquid mixture is spilled from the storage  tank it will accumulate over the floor space.

Thus we infer

$V_{spilled}=V_{accumulated}$

where

$V_{spilled}$ is spilled volume of mixture = $2900m^{3}$

$V_{accumulated}$ is accumulated volume over the floor

Since the floor is smooth thus we conclude that the accumulated volume will form a prism of liquid with height 'h'

Thus  

$V_{accumulated}=Area\times h=40000\times h$

Equating both the volumes we get

$40000\times h=2900m^{3}\times \frac{35.314ft^{3}}{m^{3}}=102412.533ft^{3}\\\\\therefore h=\frac{102412.533}{40000}=2.56 feet$

Thus the height formed by the liquid equals 2.56 feet.

Part b)

We know that the basic relation between density, mass and volume is

$Mass=density\times volume$

Since it is given that the density of the liquid is 112% the density of water thus we have

$density_{liquid}=1.12\times density_{water}\\\\density_{liquid}=1.12\times 1000=1120kg/m^3$

Thus the weight of the volume of liquid is

$mass=2900\times 1120=3248000kg$

Now since it is given that 1 truck can carry 9000 kg of liquid thus by proportion the number of trucks need to carry 3248000 kilograms of liquid equals

$n=\frac{3248000}{9000}=360.88\approx 361$