A blacksmith heated a metal whose cubic expansivity is 6.3^-6k-1. The area expansivity of the metal is …….
if s is side of cube
and ds/s /dT = c (coef of linear exp)
ds /s= c dT
original area= s^2
new area = (s+ds)^2 = s^2 + 2 s ds + ds^2
increase in area = 2 s ds+ds^2
for small ds increase = 2 s ds
so increase/s^2 = 2 ds/s = 2 c dT
square coef = 2 *linear coef
original vol =s^3
final vol = (s + ds)^3 = s^3 + 3 s^2 ds + s ds^2 + ds^3
for small ds
change = 3 s^2 ds
change/s^3 = 3 ds /s = 3 c dT
so 3 times the linear and twice the area
so
area coef = 2/3 cubic coef
To find the area expansivity of the metal, we can use the relationship between cubic expansivity (β) and area expansivity (α):
β = 2α
Given that the cubic expansivity (β) of the metal is 6.3 x 10^-6 k^-1, we can substitute this value into the equation:
6.3 x 10^-6 k^-1 = 2α
Now, let's solve for α by rearranging the equation:
α = (6.3 x 10^-6 k^-1) / 2
Simplifying the expression:
α = 3.15 x 10^-6 k^-1
Therefore, the area expansivity of the metal is 3.15 x 10^-6 k^-1.
To find the area expansivity of the metal, we need to know the relationship between the linear expansivity and the area expansivity.
The linear expansivity (α) is defined as the fractional change in length (or volume) per degree Celsius change in temperature. The area expansivity (β) is defined as the fractional change in area per degree Celsius change in temperature.
The relationship between linear expansivity and area expansivity is given by the equation:
β = 2α
Therefore, to find the area expansivity (β), we need to double the value of the linear expansivity (α) given.
In this case, the linear expansivity (α) is given as 6.3^-6k-1. To find the area expansivity (β), we need to double this value.
β = 2 * 6.3^-6k-1
Simplifying further,
β = 2 * (1 / 6.3^6k)
β = 2 / 6.3^6k
The final answer for the area expansivity of the metal is 2 / 6.3^6k.