Answer:
The change in volume is [tex]\Delta V = 0.0001 \ m^3[/tex]
Explanation:
From the question we are told that
The coefficient of linear expansion is [tex]\alpha = 17 *10^{-6} \ K^{-1}[/tex]
The width of the block is [tex]b = 30 \ cm = 0.3 \ m[/tex]
The length is [tex]l = 45 \ cm = 0.45 \ m[/tex]
The thickness is [tex]h = 10 \ cm = 0.1 \ m[/tex]
The initial temperature is [tex]T_1 = 0^oC[/tex]
The final temperature is [tex]T_f = 100 ^oC[/tex]
The initial volume is mathematically represented as
[tex]V = l*b*h[/tex]
substituting values
[tex]V = 0.30 * 0.45 * 0.10[/tex]
[tex]V = 0.0135 \ m^3[/tex]
Generally the expansion equation is mathematically represented as
[tex]l' = l (1 + \alpha \Delta T)[/tex]
where [tex]l'[/tex] is the new length
substituting values
[tex]l' = 0.45 (1 + 17*10^{-6} * (100-0))[/tex]
[tex]l' = 0.4508 \ m[/tex]
The new width is evaluated as
[tex]b' = b(1 + \alpha \Delta T )[/tex]
substituting values
[tex]b'=0.30 ( 1 + 17*10^{-6} * (100 - 0))[/tex]
[tex]b'= 0.3005 \ m[/tex]
The new thickness is
[tex]h' = h(1 + \alpha \Delta T )[/tex]
substituting values
[tex]h' = 0.10 (1 + (17*10^{-6}) (100 - 0) )[/tex]
[tex]h' = 0.1001 \ m[/tex]
The new volume is mathematically evaluated as
[tex]V' = l'*b'* h'[/tex]
substituting values
[tex]V' = 0.4508 * 0.3005 * 0.1001[/tex]
[tex]V' = 0.0136[/tex]
Therefore
[tex]\Delta V = 0.0136 - 0.0135[/tex]
[tex]\Delta V = 0.0001 \ m^3[/tex]
