A copper piece weighing 7.1 kg is completely immersed in a tank filled to the top with kerosene.

A copper piece weighing 7.1 kg is completely immersed in a tank filled to the top with kerosene. What is the mass of the spilled kerosene?

Given:

m = 7.1 kilograms – the mass of the copper part, which was lowered into a tank filled with kerosene;

ro = 8900 kg / m3 (kilogram per cubic meter) – copper density;

ro1 = 800 kg / m3 – density of kerosene.

It is required to determine m1 (kilogram) – the mass of the spilled kerosene.

Let’s find the volume that the copper part occupies:

V = m / ro = 7.1 / 8900 = 0.0008 m3 (cubic meters).

Since, according to the condition of the problem, the part was completely immersed in kerosene, then, according to Archimedes’ law, the volume of the spilled kerosene will be equal to the volume of the part, that is:

V1 = V = 0.0008 m3.

Then, the mass of kerosene will be equal to:

m = V1 * ro1 = 0.0008 * 800 = 0.64 kilograms.

Answer: the mass of the spilled kerosene is 0.64 kilograms.