Please scroll down to get to the study materials.
On heating materials, temperature rises along with their length, breadth, and thickness. The expansion in the length of an object is called linear expansion while the expansion in length and breadth, i.e. in the area is called the superficial expansion. The expansion in three dimensions i.e. in length, breadth, and thickness of an object is called the cubical expansion.
Let us consider a rod of length l_{1} at θ_{1}^{0}C temperature. If we increase the temperature of the rod by heating to θ_{2}^{0}C, it will expand. Let l_{2} e the length of the rod at θ_{2}^{0}C. The increase in the length is given by
$$\Delta l= l_2-l_1 $$
Experimentally,
Change in length is found to be
(i) . Directly proportional to the original length of the body
$$\Delta l \propto l_1 \dots(i)$$
(ii). Directly proportional to increase in temperature
$$\Delta l \propto l_1 (\theta_2 -\theta_1) \dots(ii)$$
Using (i) and (ii)
$$\Delta l = \alpha l_1 (\theta_2 -\theta_1) $$
$$ \alpha=\frac{\Delta l}{l_1(\theta_2 -\theta_1)}\dots(iv) $$
If \(l_1 = 1\), l_{1}\( (\theta_2 -\theta_1) = 1\), then, \( \alpha=\Delta l\)
Hence, the coefficient of linear expansion is defined as the change in length per unit original length per unit change in temperature.
Now,
Final length = \(l_1 + \Delta l\)
$$=l_1 + l_1 \alpha(\theta_2 -\theta_1) $$
$$=l_1 (1+\alpha(\theta_2 -\theta_1)) \dots(iv) $$, which is the equation for final length of rod after change in temperature.
Let us consider metal sheet of area A_{1} at θ_{1}^{0}C temperature. If we increase the temperature of the sheet by heating to θ_{2}^{0}C, it will expand. Let A_{2} be the area of the metal sheet at θ_{2}^{0}C. The increase in the area is given by
_{$$ \Delta A = (A_2 - A_1) $$}
Experimentally,
Change in area is found to be
(i). Directly proportional to the original area of the body
$$ \Delta A\propto A_1 \dots (i)$$
(ii). Directly proportional to increase in temperature
$$\Delta A \propto (\theta_2 -\theta_1) \dots(ii)$$
Using (i) and (ii)
$$\Delta A= \beta (\theta_2 -\theta_1) $$
$$ \beta =\frac{\Delta A}{A_1 (\theta_2 -\theta_1)}\dots(iv) $$
If \(A_1 = 1\), \( (\theta_2 -\theta_1) = 1\), then, \( \beta=\Delta A\)
Hence, the coefficient of superficial expansion is defined as the change in area per unit original area per unit change in temperature.
Now,
Final Area = \(A_1 + \Delta A\)
$$=A_1 + A_1\beta(\theta_2 -\theta_1) $$
$$=A_1 (1+ \beta(\theta_2 -\theta_1)) \dots(iv) $$, which is the equation for final area of metal sheet after change in temperature.
Let us consider solid metal cube of volume V_{1} at θ_{1}^{0}C temperature. If we increase the temperature of the solid metal cube by heating to θ_{2}^{0}C, it will expand. Let V_{2} be the volume of the solid metal cube at θ_{2}^{0}C. The increase in the volume is given by
$$ \Delta V= (V_2 - V_1) $$
Experimentally,
Change in volume is found to be
(i) . Directly proportional to the original volume of the body
$$ \Delta V\propto V_1 \dots (i)$$
(ii). Directly proportional to increase in temperature
$$\Delta V \propto V_1(\theta_2 -\theta_1) \dots(ii)$$
Using (i) and (ii)
$$\Delta V= \gamma (\theta_2 -\theta_1) $$
If \(V_1 = 1\), \( (\theta_2 -\theta_1) = 1\), then, \( \gamma=\Delta V\)
Hence, the coefficient of superficial expansion is defined as the change in volume per unit original volume per unit change in temperature.
Now,
Final Volume = \(V_1 + \Delta V\)
$$=V_1 + V_1\gamma(\theta_2 -\theta_1) $$
$$=V_1 (1+ \gamma(\theta_2 -\theta_1)) \dots(iv) $$
, which is the equation for the final volume of solid metal cube after the change in temperature.
Consider a square metal sheet of a side at the temperature as shown in the figure. On heating, let its side expand to at_{.}
$$l_2 =l_1(1 + \alpha (\theta_2 -\theta_1))\dots (i) $$
If A_{2 }is area of the solid at then
$$A_2 = (l_2)^2 = (l_1(1+ \alpha\Delta\theta))^2$$
$$= l_1^2(1 +2\alpha\Delta\theta + \alpha^2\Delta\theta^2) $$
$$= A_1(1 + 2\alpha\Delta\theta + \alpha^2\delta\theta^2)$$
Where A_{1} = l_{1}^{2}, area of the plane sheet at temperature. As the value of is small,α ^{2} will be very small which can be neglected. Then,
$$A_2= A_1(1+ 2\alpha\Delta\theta) \dots (ii)$$
Again, from the superficial expansion of the sheet, we have
$$A_2= A_1(1+ \beta\Delta\theta) \dots (iii)$$
On comparing equations (ii) and (iii)
$$\beta = 2\alpha$$
$$∝=\frac{β}{2}\dots(iv)$$----------(iv)
Thus, the coefficient of superficial expansion of a sheet is twice the coefficient of linear expansion.
Consider a metal cube of a side at the temperature as shown in the figure. On heating, let its side expand to at_{.}
_{$$l_2 =l_1(1 + \alpha (\theta_2 -\theta_1))\dots (i) $$}
If V_{2 }is volume of the cube at then
$$V_2 = (l_2)^3 = (l_1(1+ \alpha \Delta \theta))^3$$
$$= l_1^3(1 +3\alpha^2\Delta\theta + \alpha \Delta \theta^2 + \alpha^3\Delta\theta^3) $$
$$= V_1^3(1 +3\alpha^2\Delta\theta + \alpha\Delta\theta^2 + \alpha^3\Delta\theta^3) $$
Where V_{1} = l_{1}^{3}, volume of the cube at temperature. As the value of is small,α ^{2} and α^{3} will be very small which can be neglected. Then,
$$V_2= V_1(1+ 3\alpha\Delta\theta)\dots (ii)$$
Again, from the superficial expansion of the sheet, we have
$$V_2= V_1(1+ \gamma\Delta\theta)\dots (iii)$$
On comparing equations (ii) and (iii)
$$\gamma = 3\alpha$$
$$\alpha=\frac{\gamma}{3}\dots(iv)$$
The relation between the three coefficients are
$$\alpha =\frac\beta 2 =\frac\gamma 3$$
When a substance in the form of wire or rod is heated, its length increases and this is known as linear expansion.
When a body is heated, the area of the body increases and it is known as superficial expansion.
When a body is heated, the volume of the body increases and it is known as cubical expansion.
.
You must login to reply
Ra
Why bottles are often easier to open dipping it in to hot water ?
Jan 18, 2017
1 Replies
Successfully Posted ...
Please Wait...
Why it hurts more in winter than in summer when our body collides with a solid object? OR tinjury caused in winter is more painful than that of summer,why?
Jan 09, 2017
0 Replies
Successfully Posted ...
Please Wait...