Question

A gold block decreasesd its density from by heating it up from 25oC to to 350oC. Use the co-effient of linear thermal expansion of gold to be 14 * 10^-6m/moC

What percentage does the density decease?

Answer 1

density=mass/volume

mass remains constant while volume increases

thermal coefficienct of volume exapnsion=3*thermal coefficient of linear expansion=42*10^(-6) m^3/(m^3.C)

let initial mass was m and volume was V.

density=D1=m/V

new volume=V*(1+thermal coefficient of volume expansion*change in temperature)

=V*(1+42*10^(-6)*(350-25))=1.0136*V

new density=D2=m/(1.0136*V)

=0.98653*D1

then percentage decrease in density=(D1-D2)*100/D1

=(D1-0.98653*D1)*100/D1

=1.347%