Given :
∆T = 57ºC – 37ºC = 20ºC
Linear expansion, α = 1.7 × 10-5 ºC-1
Cubical expansion, r = 3α
= 3 × 1.7 × 10-5
= 5.1 × 10-5 K-1
Let V be the volume of cavity, due to increase in temperature ∆T, volume increased by
∆V, ∆V = γV∆T
Or \(\frac{∆V}{V}\) = γ∆T
Thermal stress produced = B × volumetric strain
= B × \(\frac{∆V}{V}\)
= B × γ∆T
Thermal stress = 140 × 109 × 1.7 × 10-5 × 20
= 14280 × 104
= 1.428 × 108 N/m2
Thermal stress is about 103 times of atmospheric pressure i.e., 1.01 × 105 N/m2 .
