Question

Gold can be hammered into extremely thin sheets called gold leaf. an architect wants to cover a 100 ftà 89 ft ceiling with gold leaf that is five-millionths of an inch thick. the density of gold is 19.32 g/cm3, and gold costs $ 953 per troy ounce (1 troy ounce = 31.1034768 g). how much will it cost the architect to buy the necessary gold?

Answer 1

Answer: $62160.534

Explanation:

If the architect needs to cover a 100 ft x 89 ft ceiling with gold leaf sheets with a thickness of $5(10)^{-6}in$, he will need  to cover a volume $V$  as follows (assuming the ceiling has a rectangular shape):

$V=(length)(height)(width)$ (1)

Where:

$length=100 ft=1200in$

$height=89 ft=1068in$

$width=5(10)^{-6}in$

Then: $V=(1200in)(1068in)(5(10)^{-6}in)=6408 in^{3}$ (2)

On the other hand, density $\rho$ is defined as the relation between the mass $m$ of an object and its volume $V$:

$\rho=\frac{m}{V}$ (3)

As we know the density of gold is $\rho=19.32 g/cm^{3}$ and the volume the architect needs to cover, we can find the mass.

$m=\rho V$ (4)

But first, we have to change the dimensions of the volume we calculated, in order to work with the same units:

$V=6408 in^{3}=105.0083cm^{3}$

Substituting in (4):

$m=(19.32 g/cm^{3})(105.0083cm^{3})$ (5)

$m=2028.76 g$ (6) This is the amount of gold keaf sheets the architect needs.

Now, if we know $31.1034768 g$ cost $ 953, we can find how much $2028.76 g$ of gold cost:

$2028.76 g \frac{\ 953}{31.1034768 g}=\ 62160.534$

Therefore, the architect will need $62160.534 to buy the necessary gold to cover the ceiling.