A standard can of soda has a volume of 355 mL. The mass of the can is 0.369 kg when it is full, while the can weighs 0.153 N when empty. Determine the density, specific weight, and specific gravity of the soda, and compare your results to the corresponding values for water.
The Correct Answer and Explanation is:
Let’s break this down step by step.
1. Density (ρ) of the Soda
Density is defined as mass per unit volume: ρ=mV\rho = \frac{m}{V}ρ=Vm
Where:
- mmm is the mass of the soda
- VVV is the volume of the soda
The total mass of the can when full is 0.369 kg. The mass of the empty can is determined from its weight: W=mgW = mgW=mg
Where:
- W=0.153 NW = 0.153 \, \text{N}W=0.153N (weight of the empty can)
- g=9.81 m/s2g = 9.81 \, \text{m/s}^2g=9.81m/s2 (acceleration due to gravity)
The mass of the empty can: mempty=Wg=0.1539.81=0.0156 kgm_{\text{empty}} = \frac{W}{g} = \frac{0.153}{9.81} = 0.0156 \, \text{kg}mempty=gW=9.810.153=0.0156kg
Now, the mass of the soda itself: msoda=mfull−mempty=0.369−0.0156=0.3534 kgm_{\text{soda}} = m_{\text{full}} – m_{\text{empty}} = 0.369 – 0.0156 = 0.3534 \, \text{kg}msoda=mfull−mempty=0.369−0.0156=0.3534kg
We are given the volume of the soda as 355 mL, which is equivalent to 0.355 L or 0.355 × 10⁻³ m³.
Now we can calculate the density: ρsoda=msodaV=0.35340.355×10−3=997.5 kg/m3\rho_{\text{soda}} = \frac{m_{\text{soda}}}{V} = \frac{0.3534}{0.355 \times 10^{-3}} = 997.5 \, \text{kg/m}^3ρsoda=Vmsoda=0.355×10−30.3534=997.5kg/m3
2. Specific Weight (γ) of the Soda
Specific weight is the weight per unit volume: γ=WV\gamma = \frac{W}{V}γ=VW
Where:
- W=msoda×g=0.3534×9.81=3.464 NW = m_{\text{soda}} \times g = 0.3534 \times 9.81 = 3.464 \, \text{N}W=msoda×g=0.3534×9.81=3.464N
Thus, the specific weight: γsoda=3.4640.355×10−3=9755.8 N/m3\gamma_{\text{soda}} = \frac{3.464}{0.355 \times 10^{-3}} = 9755.8 \, \text{N/m}^3γsoda=0.355×10−33.464=9755.8N/m3
3. Specific Gravity (SG) of the Soda
Specific gravity is the ratio of the density of the substance to the density of water at the same temperature. The density of water is approximately 1000 kg/m³ at 4°C. SG=ρsodaρwater=997.51000=0.9975SG = \frac{\rho_{\text{soda}}}{\rho_{\text{water}}} = \frac{997.5}{1000} = 0.9975SG=ρwaterρsoda=1000997.5=0.9975
Comparison to Water:
- Density of Water: 1000 kg/m31000 \, \text{kg/m}^31000kg/m3
- Specific Weight of Water: 9800 N/m39800 \, \text{N/m}^39800N/m3
- Specific Gravity of Water: 111
Conclusion:
- The density of the soda is very close to that of water (997.5 kg/m³ vs. 1000 kg/m³), indicating that the soda has a similar density to water.
- The specific weight of the soda (9755.8 N/m³) is slightly less than that of water (9800 N/m³), which again shows that the soda is nearly the same as water in weight per unit volume.
- The specific gravity is about 0.9975, almost identical to that of water (SG = 1).
This comparison shows that soda is nearly as dense and heavy as water, but with a slightly lower density and specific weight.
