Two pendulum clocks, one with an iron pendulum ( α = 12 × 1 0 − 6 per °C) and one with brass ( α = 19 × 1 0 − 6 per °C), agree at 10°C. How many seconds per day do they differ at 50°C? (one decimal)

T ∝ L , so fractional change is 2 1 α Δ T . Difference per day = 2 1 ( 7 × 1 0 − 6 ) ( 40 ) ( 86400 ) = 12.1 s .

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PhysicsHeat & ThermoThermal ExpansionEasy2 minQB

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Two pendulum clocks, one with an iron pendulum ( $α = 12 \times 1 0^{- 6}$ per °C) and one with brass ( $α =$ $19 \times 1 0^{- 6}$ per °C), agree at 10°C. How many seconds per day do they differ at 50°C? (one decimal)

Answer:

12.10

Solution:

$T \propto L$ , so fractional change is $\frac{1}{2} α Δ T$ . Difference per day $= \frac{1}{2} (7 \times 1 0^{- 6}) (40) (86400) = 12.1 s$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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