# Assuming that the crown of King Hieron weighs 20 N in air, and 18.75 N in water, calculate the density

**Assuming that the crown of King Hieron weighs 20 N in air, and 18.75 N in water, calculate the density of the corona matter. Assuming that only silver was mixed with gold, determine how much gold was in the crown and how much silver. When solving the problem, consider the density of gold equal to 20,000 kg / m3, the density of silver – 10,000 kg / m3. What would be the volume of a solid gold crown?**

Given: Air = 20 N; Rvode = 18.75 N Find: p crown

Let’s do this task step by step, and with some explanations.

According to Archimedes’ law, the volume of the crown is equal to the volume of water weighing:

p = air – water lines = 1.25 N, that is, mass: m = P / g = 1.25 / 9.8 = 0.128 kg

Now, knowing the density of water and mass, we find the volume of the corona:

V = m / p = 0.128 / 1000 = 0.000128 m3

The mass of the corona is: m = P / g = 20 / 9.8 = 2.04 kg By the next action we find the density of the corona substance: pcrown = m / V = 15937 kg / m3

To determine how much gold and silver there were, we must compare the equation, taking the proportion of silver in the alloy as X:

10000-X + 20000 (1-X) = 15937 X = 0.4

This means that there was 40% silver in this crown, and gold – the remaining 60%.

To determine the volume of a solid gold crown, divide the mass by the density:

V = m / g = 2.04 / 2000 = 0.0001 m3