Question

What is the volume of a aluminum ball at 10°C if its volume at 160°C is 300cm^3

Answer 1

Answer:

ΔV = β (V1)(ΔT)

ΔV = β (V1)(T2 – T1)

0.5 cm3 = (72 x 10-6 oC-1)(30 cm3)(T2 – 30oC)

0.5 = (2160 x 10-6)(T2 – 30)

0.5 = (2.160 x 10-3)(T2 – 30)

0.5 = (2.160 x 10-3)(T2 – 30)

0.5 / (2.160 x 10-3) = T2 – 30

0.23 x 103 = T2 – 30

0.23 x 1000 = T2 – 30

230 = T2 – 30

230 + 30 = T2

T2 = 260oC

Answer 2

The volume of the aluminium ball at 10°C is 298.88 cm³.

Given:

Volume at $160^{o}C$ $=300cm^{3}$

To find:

Volume of the ball at $10^{o}C$

Explanation:

Lets assume an object of volume $V_{1}$ at an initial temperature $T_{1}$ and while increasing temperature to $T_{2}$ the volume of the box increases to $V_{2}$. For this,  the coefficient of expansion $(\alpha )$ of the volume is given by

$\frac{dV}{V} =\alpha .dT$

Th volumetric expansion is visible in the expansion of railway tracks in summers. To avoid damage due to the increase in the volume, some gaps are left between two adjoining tracks that would fill the gap.

Solution:

Now, according to the question, we have

$T_{1} =10^{o}C$

$T_{2}=160^{o}C$

$V_{2} =300cm^{3}$

We know, the coefficient of volume expansion $\alpha$ is

$\alpha =25$ × $10^{-6} /^{o} C$

Substituting the known values in the equation, we get

$\frac{V_{2}- V_{1} }{V_{1} } =\alpha (T_{2}- T_{1} )$

$\frac{300-V_{1} }{V_{1} } =(25*10^{-6} )(160-10)$

$\frac{300}{V_{1} } -1=0.003750$

$\frac{300}{V_{1} } =1.003750$

$V_{1} =298.879 cm^{3}$

Final answer:

Hence, the volume of the aluminium ball at 10°C is 298.88 cm³.