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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 t_{1}, the volume be V_{1} and density be D_{1}; while at temperature t_{2}, its volume becomes V_{2} and density becomes D_{2}.

∴ D_{1} = \(\frac{m}{V_1}\) and D_{2} = \(\frac{m}{V_2}\) ∴ \(\frac{D_1}{D_2}\) = \(\frac{V_2}{V_1}\)

If the coefficient of volume expansion of the solid is γ, then

[neglecting higher powers of γ(t_{2} – t_{1}), as γ is very small]

If D_{0} and D_{t} are the densities of the solid at 0 and t degree temperatures respectively, equation (2) reduces to D_{t} = D_{0}(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 D_{1} = D_{2} [1 + γ(t_{2} – t_{1})], and substituting the given values,

D_{1} = 2.6 g ᐧ cm^{-3}, D_{2} = 2.596 g ᐧ cm^{-3}, t_{1} = 10°C and t_{2} = 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 P_{1} and P_{2} 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 P_{1} and the rod B is heated. When the rod expands, the pin at P_{1} 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 P_{1}.

Expansion of the rod, when obstructed, also generates a huge force. To demonstrate this, a pin is inserted through P_{2} and the rod is fixed rigidly by adjusting the screw N. If now the rod is heated, the pin P_{2} 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 cm^{2}. 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 × 10^{12} dyn ᐧ cm^{-2} and its coefficient of linear expansion is 12 × 10^{-6}°C^{-1}]

**Solution:**

In this case, A = 4 cm^{2}, Y = 2.1 × 10^{12} dyn ᐧ cm^{-2}, α = 12 × 10^{-6}°C^{-1} and t = 60 – 30 = 30°C

∴ The force exerted

= AYαt =4 × 2.1 × 10^{12} × 12 × 10^{-6} × 30

= 3.024 × 10^{9} 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 cm^{2}; α = 16 × 10^{-6}°C^{-1}, Y = 20 × 10^{11} dyn ᐧ cm^{-2}

**Solution:**

Here A = 0.01 cm^{2}, Y = 20 × 10^{11} dyn ᐧ cm^{-2}, α = 16 × 10°C^{-1}, t = 10°C

∴ Change in tension

AYαt = 0.01 × 20 × 10^{11} × 16 × 10^{-6} × 10

= 32 × 10^{5} dyn.