Answer
The specific gravity of the wood is 0.88
Work Step by Step
Let $M_a$ be the apparent mass of the wood. Let $M$ be the true mass of the wood. Let $\rho_a$ be the density of alcohol. We can find the volume $V$ of the wood. The apparent weight is equal to the actual weight minus the buoyant force.
$M_a~g = M~g-F_B$
$F_B = M~g-M_a~g$
$\rho_a~V~g = M~g-M_a~g$
$V = \frac{M-M_a}{\rho_a}$
$V = \frac{0.48~kg-0.047~kg}{0.79\times 10^3~kg/m^3}$
$V = 5.48\times 10^{-4}~m^3$
We can find the density $\rho$ of the wood as:
$\rho = \frac{0.48~kg}{5.48\times 10^{-4}~m^3}$
$\rho = 0.88\times 10^3~kg/m^3$
The specific gravity is the ratio of the density of the material to the density of water when water has a density of $1.00\times 10^3~kg/m^3$. Since the density of the wood is $0.88\times 10^3~kg/m^3$, the specific gravity of the wood is 0.88.
