a) Density at 100 degrees: [tex]1.34\cdot 10^4 kg/m^3[/tex]
Explanation:
The density of mercury at 0 degrees is [tex]d=1.36\cdot 10^4 kg/m^3[/tex]
Let's take 1 kg of mercury. Its volume at 0 degrees is
[tex]V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3[/tex]
The formula to calculate the volumetric expansion of the mercury is:
[tex]\Delta V= \alpha V \Delta T[/tex]
where
[tex]\alpha=180\cdot 10^{-6} K^{-1}[/tex] is the cubic expansivity of mercury
V is the initial volume
[tex]\Delta T[/tex] is the increase in temperature
In this part of the problem, [tex]\Delta T=100 C-0 C=100 C=100 K[/tex]
So, the expansion is
[tex]\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3[/tex]
So, the new density is
[tex]d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3[/tex]
b) Density at 22 degrees: [tex]1.355\cdot 10^4 kg/m^3[/tex]
We can apply the same formula we used before, the only difference here is that the increase in temperature is
[tex]\Delta T=22 C-0 C=22 C=22 K[/tex]
And the volumetric expansion is
[tex]\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3[/tex]
So, the new density is
[tex]d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3[/tex]
