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THERMAL PROPERTIES OF MATTER 207 much as order(s) of magnitude larger than the Answer ∆A 3 (∆a) (∆b) coefficient of volume expansion of typical liquids. There is a simple relation between the ∆A = a (∆b) coefficient of volume expansion ( α ) and l v coefficient of linear expansion ( α l. Imagine a ∆b cube of length, l, that expands equally in all directions, when its temperature increases by b ∆T. We have a ∆a ∆l =αll ∆T so, ∆V = (l+∆l) – l≃ 3l ∆l (10.7) 2 3 In Equation (10.7), terms in (∆l) and (∆l) have been neglected since ∆l is small compared to l. So ∆A 2 b (∆a) 3V ∆l Fig. 10.8 ∆V = =3Vα ∆l (10.8) l Consider a rectangular sheet of the solid which gives material of length a and breadth b (Fig. 10.8 ). α = 3α (10.9) When the temperature increases by ∆T, a v l increases by ∆a = α a∆T and b increases by ∆b What happens by preventing the thermal l = αlb ∆T. From Fig. 10.8, the increase in area expansion of a rod by fixing its ends rigidly? ∆A = ∆A +1A + ∆2 3 Clearly, the rod acquires a compressive strain ∆A = a ∆b + b ∆a + (∆a) (∆b) due to the external forces provided by the rigid 2 2 = a αl ∆T + b αla ∆T + (α l ab (∆T) support at the ends. The corresponding stress = αlab ∆T (2 + αl∆T) = αlA ∆T (2 + αl∆T) set up in the rod is called thermal stress. For –5 –1 example, consider a steel rail of length 5 m and Since αl ≃ 10 K , from Table 10.1, the 2 product α lT for fractional temperature is small area of cross-section 40 cm that is prevented in comparision with 2 and may be neglected. from expanding while the temperature rises by 10°C. The coefficient of linear expansion of steel Hence, –5 –1 isα l(steel)2 × 10 K . Thus, the compressive ⊳ ∆l –5 –4 strain is l = αl(steel) 1.2 ×10 × 10=1.2× 10 . 11 –2 ⊳Example 10.2 A blacksmith fixes iron ring Youngs modulus of steel is Y (steel) ×10 N m . Therefore, the thermal stress developed is on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ∆F ∆l  ring are 5.243 m and 5.231 m, respectively = Ystee = 2.4 × 10 7 N m , which A  l  at 27 °C. To what temperature should the corresponds to an external force of ring be heated so as to fit the rim of the wheel?  ∆l  7 –4 5 ∆F = AY stell = 2.4×10 ×40 ×10 j 10 N. If Answer two such steel rails, fixed at their outer ends, are in contact at their inner ends, a force of this Given, T1= 27 °C magnitude can easily bend the rails. LT1= 5.231 m L = 5.243 m ⊳ T2 Example 10.1 Show that the coefficient of So, area expansion, (∆A/A)/∆T, of a L =L [1+α (T –T )] rectangular sheet of the solid is twice its T2 T1 l 2 1 5.243 m = 5.231 m [1 + 1.20×10 K (T –27°C)] linear expansivity, αl. 2 or T = 218 °C. ⊳ 2 Reprint 2026-27