Question

An iceberg has a volume of 7685 cubic feet. What is the mass of the ice (in kg) composing the iceberg?

Answer 1

Volume of an iceberg = 7685 cubic feet

Since, density of ice = 0.917 $g/cm^3$

We have to determine the mass of the ice.

Since density = $mass \times volume$

Since the unit of density is in $g/cm^3$, first we will convert the density unit in kilogram.

Since, 1 kilogram = 1000 gram

Therefore, density = $\frac{0.917}{1000} kg/cm^3$

= 0.000917 $kg/cm^3$

Since, the density unit is in cubic centimeter. We will convert it into cubic feet.

1 cubic centimeter = 0.00003531466 cubic feet

Therefore, 0.000917  $kg/cm^3$ = 0.000917 x 0.00003531466

= 0.0000000323 $kg/ft^3$.

Now, mass = $density \times volume$

= 0.0000000323 $\times$ 7685

= 0.000248

=0.00025 $kg$.

Therefore, the mass of ice is 0.00025 $kg$.

Answer 2

Answer:

199,552.74 kg

Step-by-step explanation:

The density of ice is 917 kg/m³. The density of an object is defined as the ratio of the mass of the object to the volume of the object.

Given that the volume of the ice is 7685 cubic feet and we known that

1 m³ = 35.3147 cubic feet

Therefore 7685 cubic feet

= 7685/35.3147

= 217.614761 m³

Since Density = mass/vloume

mass = density * volume

= 917 kg/m³ * 217.614761 m³

= 199,552.7358

≈ 199,552.74 kg