P 64 Ch 18 Halliday and Resnick 7th ed poses the problem of an aluminum cube mea
ID: 2096482 • Letter: P
Question
P 64 Ch 18 Halliday and Resnick 7th ed poses the problem of an aluminum cube measuring .02m on edge floating in mercury. The temp of the mercury is raised 50K. The coefficient of volume expansion of Hg is given as 1.8 X 10*-4/ K. I don't follow the solution in Chegg which arrives at an answer of 2.7 X 10*-4 cm for the change in the hight of the mercury on the edge of the cube.
My approach is to note that the mass of the cube is easitly found from the density and volume(given) of Al. This will be the same as the mass of dislacement of Hg from which the differential vol of displacement, dV, of Hg can be determined from the densities of Hg at the two temperatures given( 270K - 320K). Then dD = dV/A where A is the area of one side of the cube.
I can't follow the solution using calculus in "Chegg answers" but I think I'm going wrong somewhere?
Explanation / Answer
volume = 0.2 * 0.2 * 0.2 m^3
= 0.008 m^3
density of alluminium = 2700 kg/m^3
so, mass = density * volume
= 0.008 * 2700
= 21.6 kg
since only mercury is heated and not the cube ,,,cross-section of cube remain the same. so the cube will sink a little in the mercury when mercury is heated(density decreased).
rho = rho0 / (1+ beta*deltaT)
= rho0 / ( 1 + 1.8 X 10^(-4) )
,,,,, volume------- proportional to------ 1 / density
area * height ------- proportional to------ density,,
since cross-section remains constant
height ------- proportional to------ density,,
so,, rho / rho0 = h0 / h
or 1 / (1 + 1.8 * 10^(-4 ) ) = h0 / h --------(1),,,,,,,
now finding h0...
h0 / 0.02 = 2700 / 13534
or h0 = 0.003989951 m
now putting this in equation(1),,
we get, h = 0.00399066 m ,,,,
change in height = h - h0
= 7.09 * 10^( - 7 ) m
= 7.09 * 10^( - 5 ) cm
there is no need for calculus in this question.. hopefully this helps
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