Thermal Energy and Expansivity "Part 2" Calculation of thermal Expansivity

Hello guys, I've come back and like I said I'll teach you how to calculate thermal expansivity.

Firstly linear expansivity;

The parameters for this expansivity are

e = difference in length
l1=initial length
l2=final length
∅1=initial temperature
∅2=final temperature
Therefore a(as in alpha,since I can't find the sign)=e/l1∅
i.e;
a=e/l1∅ ------(eqi)
a=l2-l1/l1 (∅2-∅1) ------(eqii)
a=e=/l1(∅2-∅1) ------(eqiii)
a=l2-l1/l1∅ ------(eqiv)

Let's solve a problems as example

Example 1
A metal of 20m is heated for 5 minutes until it expands to 40m at 50°c calculate the linear expansivity of the metal.

Solution

e=40-20=20
l1=20
∅=50

Therefore a=20÷20×50=50

Secondly area expansivity

The paremeters used are

∆=the change in length
∆1=the initial lengt
∆2=the final length
l1=the initial length
∅1=initial temperature
∅2=final length
B=∆/l1∅ -----(eqv)
B=∆/l1(∅2-∅1) -----(eqvi)
B=∆2-∆1/l1(∅2-∅1) ------(eqvii)
B=∆2-∆1/l1∅ -----(eqviii)
Note;B=beta as I can't find the representative sign

Let's solve a problem as example

Example 2
A steel rod of 40m is heated to 80m taking 20°c as the initial temperature calculate the area expansivity if the final length is 50°c.

Solution

B=∆2-∆1/l1(∅2-∅1)
:•80-40/40(50-20)
=40/40(30)
=30

Finally cubic expansivity

The parameters used are

μ=change in length
μ1=initial length
μ2=final length
l1=initial length
∅1=initial temperature
∅2=final temperature
r=μ/l1∅ -----(eqix)
r=μ2-μ1/l1(∅2-∅1)-----(eqx)
r=μ2-μ1/l1∅-----(eqxi)
r=μ/l1(∅2-∅1)-----(eqxii)
Note; r represent garma as I couldn't find the representative sign.

Let's solve a problem as example

Example 3
A metal rod of 20°c and length 60m is heated till the length changes to 150m
and at this time the temperature is 60°c making the difference in temperature be 40°c find the cubic expansivity

Solution

r=μ2-μ1/l1∅
:• r=150-60/60×40
r=110

With this I'm sure you can calculate Thermal Expansivity.
Thanks for stopping by, ❤️❤️❤️