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Studying Physics Topics can lead to exciting new discoveries and technological advancements.
Change of Density of A Solid due to Change of Temperature
It is known that the density of a substance \(=\frac{\text { mass }}{\text { volume }}\). With the change in temperature, while the mass of a solid remains the same, its volume changes. Hence, with the change in temperature, the density of a solid changes. With the rise in temperature, volume increases, thus density decreases; and with the decrease in temperature, volume decreases, thus density increases.
Let for a solid of mass m at temperature t1, the volume be V1 and density be D1; while at temperature t2, its volume becomes V2 and density becomes D2.
∴ D1 = \(\frac{m}{V_1}\) and D2 = \(\frac{m}{V_2}\) ∴ \(\frac{D_1}{D_2}\) = \(\frac{V_2}{V_1}\)
If the coefficient of volume expansion of the solid is γ, then
[neglecting higher powers of γ(t2 – t1), as γ is very small]
If D0 and Dt are the densities of the solid at 0 and t degree temperatures respectively, equation (2) reduces to Dt = D0(1 – γt).
Numerical Examples
Example 1.
Density of glass at 10°C is 2.6 g ᐧ cm-3 and that at 60°C is 2.596 g ᐧ cm-3. What is the average value of the coefficient of linear expansion of glass between these two temperatures?
Solution:
Using the equation D1 = D2 [1 + γ(t2 – t1)], and substituting the given values,
D1 = 2.6 g ᐧ cm-3, D2 = 2.596 g ᐧ cm-3, t1 = 10°C and t2 = 60°C,
we get, 2.6 = 2.596 [1 + γ(60 – 10)]
or, 1 + 50γ = \(\frac{2.6}{2.596}\) or, 50γ = \(\frac{2.6-2.596}{2.596}\)
or, 50γ = 1.00154 – 1
or, γ = \(\frac{0.00154}{50}\) = 30.8 × 10-6°C-1
∴ α = \(\frac{\gamma}{3}\) = 10.27 × 10-6°C-1
Thermal Stress
A change in temperature causes a change in the length of a metal rod. But if the two ends of the rod are rigidly fixed at fixed supports, expansion or contraction of the rod gets obstructed. Hence, a large force is generated along the rod. This force, measured per unit area of the rod is called thermal stress.
Experimental demonstration of thermal stress: A metal rod B is set within the gap of a heavy iron frame Y [Fig.]. One end of the rod B is threaded and two holes P1 and P2 are at the other end. A screw N is fitted at the threaded end of B. A cast iron pin is introduced through the hole P1 and the rod B is heated. When the rod expands, the pin at P1 is tightly fitted with the frame by adjusting the screw N. If the rod is cooled now, the pin obstructs the contraction of the rod developing a huge force which breaks the pin inserted through P1.
Expansion of the rod, when obstructed, also generates a huge force. To demonstrate this, a pin is inserted through P2 and the rod is fixed rigidly by adjusting the screw N. If now the rod is heated, the pin P2 gets broken due to the force developed in the rod on expansion.
Magnitude of thermal stress: Let a rod of length l, cross-sectional area A, coefficient of linear expansion α be heated so that the rise in temperature is t. The rod, therefore expands by lαt. Now if the two ends of the rod are rigidly fixed and it is cooled to its original temperature, it tends to contract back to its original length and this contraction is opposed by a force F (say).
Therefore, the reaction to the force F, which is equal and opposite to F, is the thermal force developed in the rod due to expansion lαt.
From Hooke’s law, the Young’s modulus of the rod,
Therefore, the thermal stress = \(\frac{F}{A}\) = Yαt
Clearly, thermal stress is independent of length or area of cross-section of the rod (or a wire).
Numerical Examples
Example 1.
Two ends of a steel rod are rigidly fixed with two supports. At 30 °C its area of cross-section is 4 cm2. How much force will be exerted on the supports by the ends of the rod if the temperature of the rod is raised by 60°C? [Young’s modulus of steel = 2.1 × 1012 dyn ᐧ cm-2 and its coefficient of linear expansion is 12 × 10-6°C-1]
Solution:
In this case, A = 4 cm2, Y = 2.1 × 1012 dyn ᐧ cm-2, α = 12 × 10-6°C-1 and t = 60 – 30 = 30°C
∴ The force exerted
= AYαt =4 × 2.1 × 1012 × 12 × 10-6 × 30
= 3.024 × 109 dyn.
Example 2.
Two ends of a wire are rigidly clamped. If its tem-perature is decreased by 10°C, find the change in the tension of the wire.
Area of cross-section of the wire = 0.01 cm2; α = 16 × 10-6°C-1, Y = 20 × 1011 dyn ᐧ cm-2
Solution:
Here A = 0.01 cm2, Y = 20 × 1011 dyn ᐧ cm-2, α = 16 × 10°C-1, t = 10°C
∴ Change in tension
AYαt = 0.01 × 20 × 1011 × 16 × 10-6 × 10
= 32 × 105 dyn.