THERMAL PROPERTIES OF MATTER

1) A sheet of copper has an area of 500cm² at 0˚C. find the area of this sheet at 80˚C.
The coefficient of area expansion is to good approximation 2α.thus

Answe:

∆A=2αA∆t
=2(1.67x10⁻⁵)(500)(80˚-0˚)
=1.34cm²

The new area is
500 cm²+1.34 cm²
=501.3cm²


2) If an anisotropic solid has coefficients of liner expansion αx, αy. Αz for three mutually perpendicular direction in the solid, what is the coefficient of volume expansion for the solid?

Answer:

Consider a cube,with edges parallel to X,Y,Z, of dimension L∘ at T=0. After a change in temperature ∆T=(T-0), the dimensions change to
Lx=L∘(1+αxT) Ly=L∘(1+αyT) Lz=L∘(1+αzT)
And the volume of the parallelepiped is
V=V∘(1+αxT)(1+αyT)(1+αzT)≈ V∘[1+(αx+αy+αz)T]
Where V∘=L∘³. Therefore, the coefficient of volume expansion is given by β≈αx+αy+αz. (note : the symbols β and γ are often used for the coefficient of value expansion.)


3) Calculate the increase in volume of 100cm³ of mercury when its temperature change from 10 to 35˚C.

Answer:

∆V=βV∘∆T. Then volume coefficient of expansion of mercury β=1.8x10⁻⁴˚C⁻¹. then ∆V=(1.8x10⁻⁴˚C⁻¹)(100cm³)(25˚C)=0.45cm³


4) The coefficient of liner expansion of glass. If a bottle holds 50.000cm³ at 15˚C, what is its capacity at 25˚C?

Answer:

The glass expands uniformly in all dimensions, so the capacity expands in proportion to the glass. The volume coefficient of expansion is related to the linear coefficient byβ=3α. Then V=V∘(1+3α∆T)=(50.000cm³)[1+3(8.3x10⁻⁶˚C⁻1)(10˚c)]. V=50.012cm³


5) Determine the change in volume of a block of iron 5cm x 10cm x 6cm, when the temperature change from 15˚C to 47˚C.

Answer:

V∘=300cm³; β=3α.then
∆V=3αV∘∆T=3(1.2 x 10⁻⁵˚C⁻¹)(300cm³)(32˚C)=0.35cm³



6) A glass vessel is filled with exactly 1L of turpentine at 20˚C. What volume of the liquid will overflow if the temperature is raised to 86˚C?

Answer:

As the temperature raises, both the capacity of the glass vessel and the volume of the turpentine increase. The amount of the overflow is the difference in these increases : ∆VT-∆VG=(βT-βG)V∆T. noting that βG=3αG, and using the data , we have
∆VT-∆VG=(9.4x10⁻⁴˚C⁻¹)-(0.25 x 10⁻⁴˚C⁻¹)(1L)(66˚C)=60mL


7) The density of gold is 19.30g/cm³ at 20˚C. compute the density of gold at 90˚C.

Answer:

d=M/V d1/d2=V2/V1
since mass constant. The indices 1 and 2 refer to 90˚C and 20˚C, respectively
d1/d2=V2/[V2(1+3α∆T)]
=1/[(1+3α∆T)]≈ 1-3α∆T (since 3α∆T<<1)
d1=d2(1-3α∆T)=(19.3g/cm³)[1-3(14.2 x 10⁻⁶˚C⁻¹)(70˚C)]
d1=19.24g/cm³


8) The density of mercury at 0˚C is 13600kg/m³.calculate the density of mercury at 50˚C.

Answer:

Let ρ1=ρ∘[V∘/V1]=ρ∘[V∘/(V∘+∆V)]=ρ∘[1/(1+(∆V/V∘)]
But ∆V/V∘=β∆T=(1.82 x 10⁻⁴˚C⁻¹)(50˚C)=0.0091
Substitution gives ρ1=(13600kg/m³)[1/(1+0.0091)]=13477kg/m³


9) the mercury in a thermometer has a volume of 210mm³ at 0˚C, at which temperature the diameter of the bore is 0.2mm. How far apart are the degree marks on the stem?

Answer:

We assume that the temperature rises from 0˚C to t. so ∆t=t. Then
∆VHg =β Hg V Hg ∆t, ∆VG = 3αGVG ∆t, where here VHg is the volume occupied by mercury at 0˚C and VG is the volume of the glass container occupied by the mercury at 0˚C. Indeed hAt = ∆VHg-∆VG, where h is the height that the mercury rises and At is the cross section of the bore at temperature t. At = ∏r(1+2αG∆t) with r=0.1mm, the radius at 0˚C, then
Sh=[(βHg-3αG)VHgt]/[ ∏r²(1+2αG t)]
=[ [(182-25) x 10⁻⁶˚C⁻1](210mm³)t]/[0.0314[1+(16.6 x 10⁻⁶˚C⁻¹)t]]


10) A glass beaker hold exactly 1L at 0˚C. (a) what is its volume at 50˚C? (b) If the beaker is filled with mercury at 0˚C, what volume of mercury overflow when the temperature is raised to 50˚C?

Answer:

a) The volume of the beaker after the temperature change is.
Vbeaker =V∘(1+3αG∆T)=(1)[1+3(8.3 x 10⁻⁶)(50)]=1.001L.
b)for the mercury expansion,
Vmercury=V∘(1+β∆T)=(1)[1+(1.82 x 10⁻⁴)(50)]=1.009L
The overflow is thus 1.009-1.001=0.008L, or 8mL