Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 13 - Temperature and Kinetic Theory - General Problems - Page 388: 77

Answer

a) Lower in mercury. b) $0.36\%$

Work Step by Step

a) We know that increasing the temperature decreases the density. And since mercury is liquid and the cubic of iron is solid, so the mercury will have a greater change in density than that of iron. So, we can say that the density of mercury decreases while the density of iron remains constant (or its change in density is too small to be mentioned). Thus, the iron cube will go deeper into the mercury. In other words, $\text{the iron cube will float lower in mercury}$. $\rightarrow$ In pure physics languages: We know that $\rho_{Hg}\gt \rho_{Fe}$ We also know that $\beta_{Hg}\gt \beta_{Fe}$ which means that the increase in temperature causes a change in densities and this change is greater in mercury as well. $\Delta \rho_{Hg}\gt \Delta\rho_{Fe} $ So that the iron cube will float lower in mercury. b) We know that both volumes of mercury $\left(V_{Hg}\right)_{displaced}$ and iron $\left(V_{Fe}\right)_{submerged}$ will expand as we raise the temperature. Thus, the initial submerged fraction of the iron is given by Thus, the fraction of volume submerged is given by $${\text{Fraction volume submerged}}=\dfrac{ \left(V_{Fe}\right)_{submerged}}{ V_{Fe} }=\dfrac{\left(V_{Hg}\right)_{displaced}}{V_{Fe} }$$ So that, $${\text{Fraction volume submerged}} =\dfrac{\left(V_{Hg}\right)_{displaced}}{V_{Fe} }\tag 1$$ Hence, the fractional change is given by $${\text{Fractional change}}=\dfrac{\dfrac{\left(V_{Hg}\right)_{displaced}}{V_{Fe} }-\dfrac{\left(V_{i,Hg}\right)_{displaced}}{V_{i,Fe} } }{\dfrac{\left(V_{i,Hg}\right)_{displaced}}{V_{i,Fe} }}\tag 2$$ Note that $i\rightarrow$ initial. Noting that te change in volume is given by $$\Delta V=V-V_i=V_i\beta \Delta T $$ Thus, the final volume is given by $$V=V_i\left(1+\beta \Delta T\right)$$ Using this formula into (2); $${\text{Fractional change}}=\dfrac{\dfrac{\left(V_{i,Hg}\right)_{displaced}\left(1+\beta_{Hg} \Delta T\right)}{V_{i,Fe}\left(1+\beta_{Fe} \Delta T\right)}-\dfrac{\left(V_{i,Hg}\right)_{displaced}}{V_{i,Fe} } }{\dfrac{\left(V_{i,Hg}\right)_{displaced}}{V_{i,Fe} }} $$ You can see that $\dfrac{\left(V_{i,Hg}\right)_{displaced}}{V_{i,Fe}}$ is a common factor. So that $${\text{Fractional change}}= \dfrac{ \left(1+\beta_{Hg} \Delta T\right)}{ \left(1+\beta_{Fe} \Delta T\right)}-1 $$ Plugging from table 13-2, and recall that $\Delta T=25^\circ\;\rm C$ $${\text{Percentage Fractional change}}=\left[ \dfrac{ \left(1+180\times10^{-6}\times 25 \right)}{ \left(1+35\times10^{-6}\times 25 \right)}-1\right] \times 100\% $$ $$\boxed{{\text{Percentage Fractional change}}=\bf 0.36\%} $$
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