Question

A hunk of aluminum is completely covered with a gold shell to form an ingot of weight 47.0 N . When you suspend the ingot from a spring balance and submerge the ingot in water, the balance reads 40.0 N .

Part A

What is the weight of the gold in the shell?

In newtons

Answer 1

The weight of the water displaced = 47 N - 40 N = 7 N
The mass of the water displaced = 7 N/9.8 m/s*2 = 0.714 kg = 714 g
Since the density of water is 1 g/cc, the volume of water displaced = 714 cc = volume of the ingot

volume(gold) + volume(aluminum) = 714 cc
m(Au)/19.3 g/cc + m(Al)/2.7 g/cc = 714 cc
0.14 m(Au) + m(Al) = 1927.8 cc

m(Au) + m(Al) = 47 N / 9.8 m/s*2 = 4795.92 g
m(Al) = 4795.92 - m(Au)

0.14m(Au) + 4795.92 - m(Au) = 1927.8
0.86 m(Au) = 2868.12
mass of gold = 3335.02 g = 3.335 kg

Hope that helped