Respuesta :
Answer:
The length after elongation is 0.1243 m
Solution:
As per the question;
Length of the iron vat, L = 14 m
Temperature, T = [tex]20^{\circ}[/tex] = 293 K
Temperature at 1 atm, T' = [tex]100^{\circ}[/tex] = 373 K
Pressure, P = 1 atm
Thermal expansion coefficient, [tex]\alpha = 11.1\times 10^{- 6}/^{\circ}C[/tex]
Now,
Change in length is given by:
[tex]\Delta L = \alpha L\Delta T[/tex]
[tex]L' - L = \alpha L\Delta T[/tex]
[tex]L' = L(1 + \alpha L\Delta T)[/tex]
[tex]L' = 14(1 + 11.1\times 10^{- 6}(100 - 80)[/tex]
L' = 0.1243 m
Explanation:
It is given that,
Length of iron bar, l = 14 m
Room temperature, T = 20° C
Pressure, P = 1 atm
The change in length due to change in temperature is called linear expansion. It is given by :
[tex]\Delta l=l\alpha \Delta T[/tex]
[tex]\alpha[/tex] is the coefficient of linear expansion,[tex]\alpha =12\times 10^{-6}\ C^{-1}[/tex]
We know that the boiling temperature of water T' = 100° C
[tex]\Delta l=l\alpha (T'-T)[/tex]
[tex]\Delta l=14\times 12\times 10^{-6}\times (100-20)[/tex]
[tex]\Delta l=0.0134\ m[/tex]
So, the new length of the iron rod is, [tex]L=14\ m+0.0134\ m=14.013\ m[/tex]
Hence, this is the required solution.
