Question

A .48kg piece of wood floats on water but is found to sink in alcohol (sg=.79). What is the sg of the wood?

Answer 1

Answer on Question #58254, Physics / Molecular Physics | Thermodynamics

A .48kg piece of wood floats on water but is found to sink in alcohol (sg=.79). What is the sg of the wood?

Solution:

Well, obviously it is between .79 and 1.

You need to know it's apparent weight or mass in alcohol $M_a$.

The apparent weight of the wood in alcohol is

$$W_a = M_a g = M g - F_b$$

Solving for the buoyant force

$$F_b = M g - M_a g = (M - M_a) g$$

According to Archimedes principle

$$F_b = \rho_a V_{df} g$$

Combining (3) and (2), we get

$$\rho_a V_{df} g = (M - M_a) g$$ $$\rho_a V_{df} = (M - M_a)$$ $$V_{df} = (M - M_a) / \rho_a$$

By definition specific gravity is the density of the substance divided by the density of water so that can write

$$s g_a = \frac{\rho_a}{\rho_w}$$

or

$$\rho_a = S g_a * \rho_w$$

Substituting, we get

$$V_{df} = \frac{M - M_a}{[s g_a \rho_w]}$$

Now, since the block of wood sinks in alcohol, it must be totally immersed in the liquid, so that the volume of displaced fluid $V_{df}$ must be equal to the volume of the block of wood $V_{bw}$

$$(V_{bw} = V_{df}).$$

It follows that

$$M = \rho_{bw} * V_w = \rho_{bw} * V_{df} = \rho_{bw} * \frac{M - M_a}{[s g_a * \rho_w]}$$ $$M = s g_w * \frac{M - M_a}{s g_a}$$

since by definition of specific gravity $SG_w = \frac{\rho_{bw}}{\rho_w}$.

Finally, from (7)

$$s g_w = \frac{M \cdot s g_a}{M - M_a}$$

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