Answer
$V=1.912$ gallons of water .
Work Step by Step
Given the measurements of gold bar are
length $l=0.15m$
height $h=0.05m$
width $w=0.05m$
volume of gold bar $V=l\times h\times w$
volume of gold bar $V=0.15m\times0.050m\times0.05m$
volume of gold bar $V=3.75\times10^{-4}m^3$
From table 11.1 density of gold $\rho=19300kg/m^3$
so from $\rho=\frac{M}{V}$
mass of gold bar $M=\rho V$
mass of gold bar $M=19300kg/m^3\times3.75\times10^{-4}m^3$
mass of gold bar $M=7.2375kg$
we need water of the same mass as gold bar
so mass of water $M=7.2375kg$
from table 11.1 density of water $\rho=1000kg/m^3 $
so from $\rho=\frac{M}{V}$
so from $V=\frac{M}{\rho}$
volume of water will be
$V=\frac{7.2375kg}{1000kg/m^3}$
volume of water will be $V=7.2375\times10^{-3}m^3$
$1m^3$ is equal to $264.172$ US gallons
so $V=7.2375\times10^{-3}m^3=7.2375\times10^{-3}\times264.172$gallons
$V=1911.945\times10^{-3} gallons$
$V=1.912 gallons$
