Work done through linear expansion

[LadyMario]

An 8000kg aluminum flagpole 100m long is heated by the sun from a temperature of 10°C to 20°C. Find the work done (in J) by the aluminum if the linear expansion coefficient is 24*10^-6 /°C. (The density of aluminum is 2.7*10³ kg/m³ and 1 atm = 1.0*10⁵ N/m²)

I know w=PΔV, and I know V at 10°C = mass/density= 8000/2.7*10³ = 2.963m³

I know how much longer the flagpole grows: ΔL=αLΔT = 24*10^-6(100)(20-10)=0.024m

But how does the volume at 20°C relate to the growth of the flagpole through linear expansion?

The answer is suppsoed to be 213 J.

 

[TSny]

Do you know the formula for volume expansion? The coefficient of volume expansion has a simple relation to the coefficient of linear expansion.

 

[LadyMario]

Do you know the formula for volume expansion? The coefficient of volume expansion has a simple relation to the coefficient of linear expansion.

Yea I found it in the textbook. We were never taught it in class, so I was trying to find another way to do it, but oh well :P