An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. The can’s initial volume at 8 °C is 3.5 × 10-4 m3. The coefficient of volume expansion for aluminum is 69 × 10-6 (C°)-1. When the can and the liquid are heated to 80 °C, 5.8 × 10-6 m3 of liquid spills over. What is the coefficient of volume expansion of the liquid?

Answer 1

Vc = Vo + a*(T-To)Vo
Vc=3.5*10^-4 + 69*10^-6*(80-8)3.5*10^-4=
3.56*10^-4 m^3 = Vol. of can at 80oC.

Vl = Vc + Vs = 3.56*10^-4 + 5.8*10^-6 =
3.618*10^-4 m^3=Vol. of liquid at 80oC.

3.618*10^-4 - 3.5*10^-4 = 1.18*10^-5 m^3
= The increase in vol. of the liquid.

a*(T-To)Vo = 1.18*10^-5 m^3
a*(72)3.5*10^-4 = 1.18*10^-5 m^3
a*0.0252 = 1.18*10^-5
a = 4.68*10^-4/oC. = Coefficient of vol.
expansion of the liquid.