- #1
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!
