A 5 m rod at 4 0 ∘ C has α = 20 × 1 0 − 6 K − 1 . At what Celsius temperature is its length 2 mm shorter?

Cooling needed = ( 5 × 20 × 1 0 − 6 ) 0.002 = 20 K . Final T = 40 − 20 = 2 0 ∘ C .

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PhysicsHeat & ThermoThermal ExpansionEasy2 minQB

PhysicsEasyinteger

A 5 m rod at $4 0^{\circ} C$ has $α$ $= 20 \times 1 0^{- 6} K^{- 1}$ . At what Celsius temperature is its length 2 mm shorter?

Answer:

Solution:

Cooling needed $= \frac{0.002}{( 5 \times 20 \times 1 0 ^{- 6} )} = 20 K$ . Final $T = 40 - 20 = 2 0^{\circ} C$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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