A rod is rigidly fixed at both ends. It is heated by 50 K. With Y = 2 × 1 0 11 Pa and α = 1 0 − 5 K − 1 , find elastic energy density in J / m 3 . Use the linear small-strain model.

u = 2 1 Y ( α Δ T ) 2 = 2 1 ( 2 × 1 0 11 ) ( 5 × 1 0 − 4 ) 2 = 25000 J / m 3 .

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermal ExpansionMedium2 minQB

PhysicsMediuminteger

A rod is rigidly fixed at both ends. It is heated by 50 K. With $Y = 2 \times 1 0^{11} Pa$ and $α$ $= 1 0^{- 5} K^{- 1}$ , find elastic energy density in $J / m^{3}$ . Use the linear small-strain model.

Answer:

25000

Solution:

$u = \frac{1}{2} Y (α Δ T)^{2} = \frac{1}{2} (2 \times 1 0^{11}) (5 \times 1 0^{- 4})^{2} = 25000 J / m^{3}$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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