How Does Temperature Affect the Density of an Object?

AI Thread Summary
The discussion focuses on how temperature affects the density of an object, specifically calculating the density of an object with a known density of 1250 kg/m3 at 10°C and a coefficient of linear expansion of 2.5 × 10−5 1/K at 25°C. The correct approach involves using the equation V(T)=(1+β(Δt))V0, leading to a density of 1248.6 kg/m3. The first equation mentioned, V=βV0ΔT, is clarified as applicable only for the change in volume, not the total volume. The distinction between these equations is crucial for accurate calculations in thermal expansion scenarios. Understanding when to use each equation is essential for solving related problems effectively.
Patricia Reid
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Homework Statement


An object has a density of 1250 kg /m3 at 10◦C and a coefficient of linear expansion of α = 2.5 × 10−5 1 /K . What is the object’s density when the temperature is 25◦C?

Homework Equations


V=βV0ΔT
V(T)=(1+β(Δt))V0

The Attempt at a Solution


How I got the answer is that I used V(T)=(1+β(Δt))V0 and substituted v=m/ρ and after some algebra gave me the correct answer of 1248.6 kg/m2

When I used the first equation I listed I didn't get the correct answer... I was just wondering what the difference is in the equations and when to use themThanks in advance!
 
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The first equation is not for the volume, it is for the change of volume. You can see this by inserting zero temperature difference (you will get zero and not the volume).
 

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