Volume Expansion:-
V₁ : Volume at T₁
V₂ : Volume at T₂       T₂ > T₁

V2 = V1 (1 + )

: Coefficiant of volume expansion.
For i so topic solids

: : = 1 : 2 : 3

As temperature increases density of a solid decreases.
d₂ =
d₂ < d₁
Note: in case of liquid contained in a container, both the liquid as well as conainer expand.
Thermal Stress (Secondary Information):-
When a rod is held between two rigid supports and the temperature is allowed to rise the rigid support prevents the rod from expanding.
This is the cause of stress produced in the rod. thermal expansion (if the rod were free)
l = l
Thermal stress = strain X Y =
= Y
Dum Question:-
Q. A liquid is inside a container. Both are heated what is relative coefficiant of expansion?
A. _r = _e - _c
Q. A Steel glass is filled with a liquid. The coeff. of volume expansion of steel and liquid are same.
Will the liquid overflow.
A. No because y_relative = 0
Q. A sphace of copper is floating in a murcury bath. If the temperature is increased. Sphece will sink deeper or rises? _copper > _mercury.
A. It risel because _relative = _ai - _Hg > 0
Illustration:-
Q. A Surveyor's zom steel tape is correct at a temp. of 20 ⁰C. The distance between two points as measured by this tap on a day when the temp. is 35 ⁰C, is 26 m. What is the true distance between the points?
_steel = 1.2 X 10^-5/ ⁰C
Ans. Let temp. rise above the correct temp. be
= 35 - 20 = 15 ⁰C
Using the relation
l₂ = l₁ (1 + )
True distance = 26(1 + 1.2 X 10^-5 X 15)
                     = 26.00468 m.
Assignments:-
Easy
E1. A light steel wire of length l and area of cross section A is hangig vertically downward with the ceiling. It cools to the room temp. (30 ⁰C) from initial 100 ⁰C. what weight should be attached to its lower end so that its length remains the same. Young's Modulus of steel is _s and coeff. of linear expansion is _s.
Solution:- Stress due to temperature change = _{s s}(100 - 30)
= _ss(100 - 30) w = 70 A_ss
E2. What will be the final temperature , when 300 gm. of ice at 0 ⁰C is mixed with 300 gm. of water at 50 ⁰C.
Lotent heat of fusion of ice = 80 cal/gm.
Solution:- Heat gained by ice = Heat gained by water
300 X 1 X (50 - T) = 150 X 80 + 150 X 1 (T - 0)
T = 6.7 ⁰C.
E 3. A steel ring of 3000 inches inside diameter at 20 ⁰C is to be heated and slipped over a brass shaft measuring 3.
Solution:- Final diameter of ring should be 3.002 inches
3.002 = 3[1 + ( - 20)]
= 75.6 ⁰ C.
E4. Water is being boiled in flat bottom kettle placed on a stove. The area of the bottom is 3000 Cm² and the thickness is 2 mm. If the amount of stream produced is 1 g/min., calculate the difference of temperature between the inner and outer surfaces of the bottom. K for the materrial of kettle is 0.5 cal/⁰C/g and latent heat of steam is 540 cal/g.
Solution:- Mass of steam = 1/60 g/s.

= 1.2 X 10^{-3   0}C.