Question

A quartz sphere is 14.0 cm in diameter. What will be its change in volume if its temperature is increased by 330 °F? (the coefficient of volume expansion of quartz is 1.50 × 10-6 °C-1)

Answer 1

D = 14 cm => r = 7 cm

The intial volume of sphere is:

V0 = 4/3 pi r^3 = 4/3 x 3.14 x 7^3 = 1436.76 cm^3

The change in its diameter would be,

delta D = D0 ( 1 + alpha delta T)

delta D = 14 ( 1 + 1.5 x 10-6 x 165.56) = 0.00348

chnage in diameter is = 0.00348 cm

D' = 14.00348 cm => R' = 7.00174 cm

V' = 4/3 pi R'^3 = 1437.83 cm^3

So the change in volume will be:

Delta V = V' - V = 1437.83 - 1436.76 = 1.07 cm^3

[ assumed intial temp to be zero, it should be given in such problems]