Question

The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the soda and compare your results with the corresponding values for water at 20 oC. Express your results in SI units.

Answer 1

Explanation:

The given data is as follows.

    Volume of pop in can, V = $355 \times 10^{-6} m^{3}$

• Mass of a full can of pop is as follows.

                          W = mg

                               = $0.369 \times 9.81$

                               = 3.6198 N

Weight of empty can, $w_{1}$ = 0.153 N

• Now, weight of pop in the can is calculated as follows.

                $w_{2} = W - w_{1}$

                           = 3.6198 - 0.153

                           = 3.467 N

• Calculate the specific weight of the liquid as follows.

         $\gamma = \frac{\text{weight of liquid}}{\text{volume of liquid}}$

                      = $\frac{3.467}{355 \times 10^{-6}}$

                      = 9766.197 $N/m^{3}$

• Density of the fluid is calculated as follows.

                $\rho = \frac{\gamma}{g}$

                          = $\frac{9766.197}{9.81}$

                          = 995.535 $kg/m^{3}$

• Now, specific gravity of the fluid is calculated as follows.

           S.G = $\frac{\text{density of liquid}}{\text{density of water}}$

                  = $\frac{\rho}{\rho_{w}}$

                  = $\frac{995.535}{1000}$

                  = 0.995

Answer 2

Answer:

$\rho=995.50\ kg.m^{-3}$

$\bar w=9765.887\ N.m^{-3}$

$s=0.9955$

Explanation:

Given:

• volume of liquid content in the can, $v_l=0.355\ L=3.55\times 10^{-4}\ L$

• mass of filled can, $m_f=0.369\ kg$

• weight of empty can, $w_c=0.153\ N$

So, mass of the empty can:

$m_c=\frac{w_c}{g}$

$m_c=\frac{0.153}{9.81}$

$m_c=0.015596\ kg$

Hence the mass of liquid(soda):

$m_l=m_f-m_c$

$m_l=0.369-0.015596$

$m_l=0.3534\ kg$

Therefore the density of liquid soda:

$\rho=\frac{m_l}{v_l}$ (as density is given as mass per unit volume of the substance)

$\rho=\frac{0.3534}{3.55\times 10^{-4}}$

$\rho=995.50\ kg.m^{-3}$

Specific weight of the liquid soda:

$\bar w=\frac{m_l.g}{v_l}=\rho.g$

$\bar w=995.5\times 9.81$

$\bar w=9765.887\ N.m^{-3}$

Specific gravity is the density of the substance to the density of water:

$s=\frac{\rho}{\rho_w}$

where:

$\rho_w=$ density of water

$s=\frac{995.5}{1000}$

$s=0.9955$