Question

A major motor company displays a die-cast model of its first automobile, made from 9.00kg of iron. To celebrate its hundredth year in business, a worker will recast the model in gold from the original dies. What mass of gold is needed to make the new model? The density of iron is 7.86 ✕ 10^3 kg/m3, and that of gold is 19.30 ✕ 10^3 kg/m3

Answer 1

Answer:

22.10 kg

Explanation:

The worker wants to recast the model from iron to gold, so the volume of the model made of iron and made of gold will be the same:

$V_i = V_g$

the volume of each model can be rewritten as the ratio between the mass of the model, m, and the density of the material, d:

$V=\frac{m}{d}$

so we can rewrite the first equation as

$\frac{m_i}{d_i}=\frac{m_g}{d_g}$

where we have

$m_i = 9.00 kg$ is the mass of the model in iron

$d_i = 7.86\cdot 10^3 kg/m^3$ is the density of iron

$m_g$ is the mass of gold needed

$d_g = 19.30\cdot 10^3 kg/m^3$ is the density of gold

Solving for $m_g$, we find

$m_g = d_g \frac{m_i}{d_i}=(19.30\cdot 10^3 kg/m^3) \frac{9.00 kg}{7.86\cdot 10^3 kg/m^3}=22.10 kg$