Question

A major motor company displays a die-cast model of its first automobile, made from 6.66 kg of iron. To celebrate its one-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? Use Table 9.3 in the book for the data needed to solve this problem. Give your answer in SI units and to the correct significant figure.

Answer 1

Explanation:

According to table,

  Mass of iron model ($m_{1}$) is given as 6.66 kg

   Density of gold, ($D_{2}$) = 19320 $kg/m^{3}$ unit

   Density of iron, ($D_{1}$) = 7850 $kg/m^{3}$ unit

Now, new model shape = old model shape

So, volume or iron model = volume of gold model

That is,        $V_{2} = V_{1}$

As relation between density and volume is as follows.

               Volume = $\frac{mass}{density}$

Therefore,

         $\frac{m_{2}}{D_{2}} = \frac{m_{1}}{D_{1}}$

or,          $m_{2} = D_{2} \times \frac{m_{1}}{D_{1}}$

            $m_{2} = 19320 \times \frac{6.66}{7850}$

             $m_{2}$ = 16.4 kg

Thus, we can conclude that mass of gold model, $m_{2}$ is 16.4 kg.