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- Homework Statement
- During drilling operations at Dome C in Antarctica, a cylindrical sample of glacial ice was extracted and investigated. The goal is to determine the approximate density of the ice and infer the depth at which it originated, using both geometric and physical principles.NEEM (North Greenland Eemian Ice Drilling) investigation, but uses different parameters and location to ensure independence.
- Diameter of the sample: $$ d = 7.2\,\text{cm} $$
- Height of the sample: $$ h = 1.8\,\text{cm} $$
- Mass of the sample: $$ m = 98.7\,\text{g} $$
- Beaker with water (no sample): $$ m_1 = 690.5\,\text{g} $$
- Beaker with floating sample: $$ m_2 = 817.0\,\text{g} $$
- Beaker with fully submerged sample (pushed down): $$ m_3 = 843.2\,\text{g} $$
- Density of water: $$ \rho_{\text{water}} = 1000\,\text{kg/m}^3 $$
- Relevant Equations
- Volume of a cylinder:
$$ V = \pi \cdot \left( \frac{d}{2} \right)^2 \cdot h $$
- Density:
$$ \rho = \frac{m}{V} $$
- Buoyant force (Archimedes):
$$ F_{\mathrm{buoyancy}} = (m_3 - m_2) \cdot g $$
- Volume via buoyancy:
$$ V = \frac{F_{\mathrm{buoyancy}}}{\rho_{\mathrm{water}} \cdot g} $$
We assume the sample is shaped like a perfect cylinder. The volume is given by:
$$V = \pi \cdot \left( \frac{7,2}{2} \right)^2 \cdot 1,8 = \pi \cdot 3,6^2 \cdot 1,8 = 73,2\,\mathrm{cm^3}$$
Then the approximate density is:
$$
\rho = \frac{98,7}{73,2} = 1,348\,\mathrm{g/cm^3} = 1348\,\mathrm{kg/m^3}
$$
Now using Archimedes' principle:
$$
F_{\mathrm{b}} = (843,2 - 817,0)\,\mathrm{g} \cdot 9,81 = 0,0262\,\mathrm{kg} \cdot 9,81 = 0,257\,\mathrm{N}
$$
Then the displaced volume of water is:
$$
V = \frac{0,257}{1000 \cdot 9,81} \approx 2,62 \times 10^{-5}\,\mathrm{m^3} = 26,2\,\mathrm{cm^3}
$$
So the density based on buoyancy becomes:
$$
\rho = \frac{98,7}{26,2} \approx 3,77\,\mathrm{g/cm^3} = 3770\,\mathrm{kg/m^3}
$$
This result is far too high and clearly unrealistic for ice. What did I do wrong in this approach?
$$V = \pi \cdot \left( \frac{7,2}{2} \right)^2 \cdot 1,8 = \pi \cdot 3,6^2 \cdot 1,8 = 73,2\,\mathrm{cm^3}$$
Then the approximate density is:
$$
\rho = \frac{98,7}{73,2} = 1,348\,\mathrm{g/cm^3} = 1348\,\mathrm{kg/m^3}
$$
Now using Archimedes' principle:
$$
F_{\mathrm{b}} = (843,2 - 817,0)\,\mathrm{g} \cdot 9,81 = 0,0262\,\mathrm{kg} \cdot 9,81 = 0,257\,\mathrm{N}
$$
Then the displaced volume of water is:
$$
V = \frac{0,257}{1000 \cdot 9,81} \approx 2,62 \times 10^{-5}\,\mathrm{m^3} = 26,2\,\mathrm{cm^3}
$$
So the density based on buoyancy becomes:
$$
\rho = \frac{98,7}{26,2} \approx 3,77\,\mathrm{g/cm^3} = 3770\,\mathrm{kg/m^3}
$$
This result is far too high and clearly unrealistic for ice. What did I do wrong in this approach?
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