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the nasa neutral buoyancy lab is a pool 12.00 meters in depth. if the temperature of the water increases from 20.00 to 30.00, what is the increase in depth due to the change in volume of the water?



Answer :

MathPhys

Answer:

0.02484 m

Explanation:

The depth of the pool increases due to thermal expansion. The change in volume is equal to the initial volume times the volumetric expansion coefficient times the change in temperature.

ΔV = V β ΔT

Assuming the cross section of the pool is constant:

AΔh = Ah β ΔT

Δh = h β ΔT

Plug in values:

Δh = (12.00 m) (207×10⁻⁶ K⁻¹) (10.00 °C)

Δh = 0.02484 m

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