Answer:
The value of specific heat is [tex]271[/tex] [tex]\frac{J}{Kg.K}[/tex] and coefficient of thermal expansion is [tex]1.29\times 10^{-5}[/tex]
Explanation:
Given:
Density of alloy [tex]\rho = 7380[/tex] [tex]\frac{kg}{m^{3} }[/tex]
For finding the specific heat of alloy we use formula of specific heat in case of solid material,
[tex]\rho C_{p} = 2 \times 10^{6}[/tex] [tex]\frac{J}{m^{3}. K }[/tex]
Where [tex]\rho =[/tex] density of alloy,
[tex]C_{p} = \frac{2 \times 10^{6} }{7380}[/tex]
[tex]C_{p}= 271[/tex] [tex]\frac{J}{Kg.K}[/tex]
For finding the coefficient of thermal expansion,
[tex]\alpha T = 0.02[/tex]
Where [tex]\alpha =[/tex] coefficient of thermal expansion
[tex]\alpha = \frac{0.02}{1540}[/tex]
[tex]\alpha = 1.29 \times 10^{-5}[/tex]
Therefore, the value of specific heat is [tex]271[/tex] [tex]\frac{J}{Kg.K}[/tex] and coefficient of thermal expansion is [tex]1.29\times 10^{-5}[/tex]
