How Does Temperature Affect the Buoyancy of an Iron Cube in Mercury?

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SUMMARY

The discussion centers on the buoyancy of an iron cube in mercury as temperature increases from 0°C to 30°C. The coefficient of volume expansion for mercury is 180 x 10^-6 °C^-1, while for iron it is 35 x 10^-6 °C^-1. As the temperature rises, the density of mercury decreases faster than that of iron, resulting in a greater percentage of the iron cube being submerged. The calculated change in the submerged volume percentage is 0.25%, while the textbook states it should be 0.44%, indicating a potential error in the calculations presented.

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Freye
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Homework Statement


An iron cube floats in a bowl of liquid mercury at 0degrees C.
a) If the temperature is raised to 30degrees C, will the cube float higher or lower in the mercury?
b)By what percent will the fraction of volume submerged change?
\beta mercury=180 X 10^-6
\betairon=35 X 10^-6



Homework Equations


\Delta V=\betaV_1\DeltaT
\rho=m/V
\rho_solid/\rho_fluid X 100 = % of submerged solid


The Attempt at a Solution


NOTE: all Greater than or equal to signs are meant as only greater than signs, I just couldn't find them in the Latex Reference

a)
Since the iron floats in the mercury, V_1mercury \geq V_1 iron, also \beta_mercury \geq \beta_iron
Therefore \DeltaV_mercury \geq \DeltaV_iron
then since \rho = m/V
therefore \Delta\rho_mercury decreases faster than \Delta\rho_iron
and since \rho_solid/\rho_fluid X 100 = % of submerged solid
Therefore a greater percentage of iron will be submerged

b)
\rho_1iron/\rho_1mercury= 7.8 X 10^3/(13.6 X 10^3)
therefore %of submerged iron_1 = 7.8/13.6 X 100 = 57.35%

\DeltaV_iron = 1050 X 10^-6 V_1iron
V_2iron - V_1iron = 1050 X 10^-6 V_1iron
V_2iron = 1.00105 miron/\rho_iron = 1.2834 X 10^-4 m_iron

\DeltaV_mercury = 5400 X 10 ^-6 V_1mercury
V_2mecury - V_1 mercury = 5400 X 10 ^-6 V_1mercury
V_2mercury = 1.0054 m_mercury/\rho_mercury = 7.3926 X 10^- 5m_mercury

\rho_1iron/\rho_1mercury = m_iron/V_1iron/m_mercury/V_1mercury
=m_iron V_1mercury/ V_1iron m_mercury
= [STRIKE]m_iron[/STRIKE] (7.3926 X 10^-5) [STRIKE]m_mercury[/STRIKE]/((1.2834 X 10^-4)[STRIKE]m_iron m_mercury[/STRIKE]
=.5760
% of submerged iron_2 = .5760 X 100 = 57.60%

\Delta % of submerged iron = 57.60 - 57.35 = 0.25%


I think my answer for part a) is logically sound, although there is likely an easier way to explain it. As for part b), the answer in my textbook is 0.44% not 0.25%, but I can't see where I have gone wrong. Someone please point out my mistake!
 
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where did you get the values of beta ?
What are it's units?
 
Ahh, sorry.

Beta is the coefficient of volume expansion for the given substance. I got the values from a table in my textbook, and the units are: Celsius degrees^-1
 

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