Question

An ornament weighing $150\,{\text{g}}$ in air weight only $76\,{\text{g}}$ in water. Assuming that some copper mixed with gold to prepare the ornament. Find the amount of copper in it. The specific gravity of gold is 20 and that of copper is 10.

Answer 1

Use the formulae for density of a substance and upward thrust acting on an object immersed in the liquid. Derive the expression for the volume of gold and copper in the ornament. Then use the application for Archimedes’ principle which states that the loss in weight of the ornament is equal to the upward thrust force acting on the ornament. Solve this equation and determine the mass of copper in the ornament.

Formulae used:
The density $\rho$ of a substance is given by
$\rho = \dfrac{m}{V}$ …… (1)
Here, $m$ is the mass of the substance and $V$ is the volume of the substance.
The upthrust force $F$ acting on an object in the liquid is given by
$F = \rho Vg$ …… (2)
Here, $\rho$ is density of the liquid, $V$ is volume of the object immersed in the liquid and $g$ is acceleration due to gravity.

Complete step by step answer:
We have given that the mass of an ornament made up of gold and copper in the air is $150\,{\text{g}}$ and the mass of the same ornament in water is $76\,{\text{g}}$.
${M_{Air}} = 150\,{\text{g}}$
$\Rightarrow{M_{Water}} = 76\,{\text{g}}$
The specific gravity of copper is $20\,{\text{g/cc}}$ and the specific gravity of copper is $10\,{\text{g/cc}}$.
${\rho _G} = 20\,{\text{g/cc}}$
$\Rightarrow{\rho _C} = 10\,{\text{g/cc}}$
We have asked to calculate the mass of copper in the ornament.Let us rewrite equation (1) for the specific gravity ${\rho _G}$ of the gold.
${\rho _G} = \dfrac{{{m_G}}}{{{V_G}}}$ …… (3)
Here, ${m_G}$ is the mass of gold in ornament and ${V_G}$ is the volume of copper in the ornament.

Let us now rewrite equation (1) for the specific gravity ${\rho _C}$ of the copper.
${\rho _C} = \dfrac{{{m_C}}}{{{V_C}}}$
$\Rightarrow {V_C} = \dfrac{{{m_C}}}{{{\rho _C}}}$
Here, ${m_C}$ is mass of copper in ornament and ${V_C}$ is volume of copper in the ornament.
The total mass of the ornament in air is the sum of the mass of gold and copper in the ornament.
${m_G} + {m_C} = 150\,{\text{g}}$
$\Rightarrow {m_G} = 150\,{\text{g}} - {m_C}$
Substitute $150\,{\text{g}} - {m_C}$ for ${m_G}$ in equation (3).
${\rho _G} = \dfrac{{150\,{\text{g}} - {m_C}}}{{{V_G}}}$
$\Rightarrow {V_G} = \dfrac{{150\,{\text{g}} - {m_C}}}{{{\rho _G}}}$

The total volume $V$ of the ornament is the sum of the volumes of gold and copper in the ornament.
$V = {V_G} + {V_C}$
According to Archimedes’ principle, the loss in weight of the ornament (difference between the weights of the ornament in air and in water) should be equal to the upthrust force acting on the ornament in the water.Thus, we can write
${M_{Air}} - {M_{Water}} = {\rho _{Water}}Vg$
Here, ${\rho _{Water}}$ is the density of water.
Substitute $150\,{\text{g}}$ for ${M_{Air}}$, $76\,{\text{g}}$ for ${M_{Water}}$, $1\,{\text{g/cc}}$ for ${\rho _{Water}}$ and ${V_G} + {V_C}$ for $V$ in the above equation.
$\left( {150\,{\text{g}}} \right)g - \left( {76\,{\text{g}}} \right)g = \left( {1\,{\text{g/cc}}} \right)\left( {{V_G} + {V_C}} \right)g$
$\Rightarrow 150 - 76 = {V_G} + {V_C}$
$\Rightarrow {V_G} + {V_C} = 74$

Substitute $\dfrac{{150\,{\text{g}} - {m_C}}}{{{\rho _G}}}$ for ${V_G}$ and $\dfrac{{{m_C}}}{{{\rho _C}}}$ for ${V_C}$ in the above equation.
$\Rightarrow \dfrac{{150\,{\text{g}} - {m_C}}}{{{\rho _G}}} + \dfrac{{{m_C}}}{{{\rho _C}}} = 74$
Substitute $20\,{\text{g/cc}}$ for ${\rho _G}$ and $10\,{\text{g/cc}}$ for ${\rho _C}$ in the above equation.
$\Rightarrow \dfrac{{150\,{\text{g}} - {m_C}}}{{20\,{\text{g/cc}}}} + \dfrac{{{m_C}}}{{10\,{\text{g/cc}}}} = 74$
$\Rightarrow \dfrac{{1500 - 10{m_C} + 20{m_C}}}{{200}} = 74$
$\Rightarrow 1500 + 10{m_C} = 14800$
$\Rightarrow 10{m_C} = 14800 - 1500$
$\Rightarrow 10{m_C} = 13300$
$\therefore {m_C} = 1330\,{\text{g}}$

Hence, the amount of copper in the ornament is $1330\,{\text{g}}$.

Note: One can solve the same question by another method. One can first write the equation for total mass of the ornament in terms of mass of gold and copper and then determine loss in weight of the ornament in water. From this calculation, one can determine the total volume of the ornament and convert it in terms of density and mass of gold and copper. Then solve these two equations of mass to determine the mass of copper in the ornament.