# Being the initial volume of the interior space of the

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being the initial volume of the interior space of the brass shell minus the initial volume of the steel ball. If the space occupied by the mercury did not change with temperature, the spillage would simply be the increase in volume of the mercury. However, the space occupied by the mercury does change with temperature. Both the brass shell and the steel ball expand. The interior volume of the brass shell expands as if it were solid brass, and this expansion provides more space for the mercury to occupy, thereby reducing the amount of spillage. The expansion of the steel ball, in contrast, takes up space that would otherwise be occupied by mercury, thereby increasing the amount of spillage. The total spillage, therefore, is DVMercury-DVBrass+ DVSteel. SOLUTIONTable 12.1 gives the coefficients of volume expansion for mercury, brass, and steel. Applying Equation 12.3 to the mercury, the brass cavity, and the steel ball, we have ()()()()()MercuryBrassSteelMercury0, MercuryBrass0, BrassSteel0, Steel16333316Spillage18210C1.6010m0.7010m12 CMercury5710C1.601VVVVTVTVTβββ= Δ− Δ+ Δ=ΔΔ+Δ⎤ ⎡=×°××°×°×±²²²²²²²²²²²²²³²²²²²²²²²²²²²´()()()()()331633630m12 CBrass3610C0.7010m12 C1.210mSteel°+×°×° =×±²²²²²²²²³²²²²²²²²´±²²²²²²²²³²²²²²²²²´41. SSMWWWREASONINGThe cavity that contains the liquid in either Pyrex thermometer expands according to Equation 12.3, ΔVg=βgV0ΔT. On the other hand, the volume of mercury expands by an amount ΔVm=βmV0ΔT, while the volume of alcohol expands by an amount ΔVa=βaV0ΔT. Therefore, the net change in volume for the mercury thermometer is ΔVm− ΔVg=(βmβg)V0ΔT
650TEMPERATURE AND HEATwhile the net change in volume for the alcohol thermometer is ΔVa− ΔVg=(βaβg)V0ΔTIn each case, this volume change is related to a movement of the liquid into a cylindrical region of the thermometer with volume πr2h, where ris the radius of the region and hthe height of the region. For the mercury thermometer, therefore, hm=(βmβg)V0ΔTπr2Similarly, for the alcohol thermometer ha=(βaβg)V0ΔTπr2These two expressions can be combined to give the ratio of the heights, ha/hmSOLUTIONTaking the values for the coefficients of volumetric expansion for methyl alcohol, Pyrex glass, and mercury from Table 12.1, we divide the two expressions for the heights of the liquids in the thermometers and find that hahm=βaβgβmβg=1200×10–6(C°)19.9×10–6(C°)1182×10–6(C°)19.9×10–6(C°)1=6.9Therefore, the degree marks are 6.9 times further aparton the alcohol thermometer than on the mercury thermometer. ______________________________________________________________________________ 42. REASONING AND SOLUTIONWhen the temperature is 0.0 °C, P= ρ0gh0, and when the temperature is 38.0 °C, P= ρgh.Equating and solving for hgives h= (ρ0/ρ)h0. ρ0/ρ= V/V0, since the mass of mercury in the tube remains constant. Thus, we have that h= (V/V0)h0. Now, ΔV= VV0= βV0ΔTor V/V0= 1 + βTherefore, h= (1 + βΔT)h0= {1 + [182 ×10−6(C°)−1](38.0 °C 0.0 °C)}(0.760 m)0.765 mwhere we have taken the value for the coefficient of volumetric expansion b
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