Ice Core Density Problem (Inspired by NEEM)

[jnuz73hbn]

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?

 

[kuruman]

- Mass of the sample: $$ m = 98.7\,\text{g} $$
- Beaker with water (no sample): $$ m_1 = 690.5\,\text{g} $$
- Beaker with floating sample: $$ m_2 = 817.0\,\text{g} $$

It is true that
Beaker with floating sample - Beaker with water (no sample) = Mass of the sample

It is also true that
817.0 - 690.5 = 126.5

It is not true that 126.5 = 98.7.

 

[jnuz73hbn]

well,

$$
F_{\mathrm{buoyancy}} = (m_3 - m_2) \cdot g = (843.2 - 817.0)\,\mathrm{g} \cdot 9.81 = 0.0262\,\mathrm{kg} \cdot 9.81 \approx 0.257\,\mathrm{N}
$$

The displaced volume of water is:

$$
V = \frac{0.257}{1000 \cdot 9.81} \approx 2.62 \times 10^{-5}\,\mathrm{m^3} = 26.2\,\mathrm{cm^3}
$$

$$
\rho = \frac{126.5\,\mathrm{g}}{26.2\,\mathrm{cm^3}} \approx 4.83\,\mathrm{g/cm^3} = 4830\,\mathrm{kg/m^3}
$$
This result is clearly unphysical —ice should have a density around 917–920 kg/m³.

 

[erobz]

Aren’t you measuring force (not mass) with the scale? When you forcibly submerge it with applied force $F$ the scale would measure $mg + F$?

 

[kuruman]

There is also a mistake here

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?

The 0.257 N that you calculated is the additional force that you must exert with your finger to keep the sample submerged. You forgot that the buoyant force before you pushed the sample down was equal to the weight of the sample, whatever that number was. When you push the sample down the buoyant force must be BF = weight of sample + 0.257 N.

 

[kuruman]

You have gotten all tangled up with intermediate numerical expressions. I recommend that you use a symbolic approach and substitute the numbers at the very end. I will get you started.
The first mass reading with the sample floating can be written as $$R_1=m_0+\rho_0 V_{\text{under}}$$ where
$m_0=$ the mass of beaker plus water;
$\rho_0=$ the density of the water in the beaker;
$V_{\text{under}} =$ the volume of the sample under water.

Note that the second term is the mass of the displaced water, also known as $\dots$?
A similar equation can be written when the sample is fully submerged. You have two equations and two unknowns, the volumes under water. Find a way to replace them with the mass of the sample, a known quantity, and the density of the sample, $\rho_s$, an unknown quantity. Then solve for the unknown in terms of the knowns.

The result is a dimensionless factor multiplying the density of water. There is no need for converting grams to kilograms or for introducing the acceleration of gravity.

 

[jnuz73hbn]

got it, thank you , now I have a good density