Posted by **Guadalupe** on Tuesday, May 13, 2014 at 12:24am.

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 5 oC is 3.5×10−4 m3. The coefﬁcient of volume expansion for aluminum is 69×10−6(Co)−1. When the can and the liquid are heated to 78 oC, 3.6×10−6m3 of liquid spills over. What is the coefﬁcient of volume expansion of the liquid?

- physics -
**MathMate**, Tuesday, May 13, 2014 at 6:51am
Initial volume of liquid, V

= 3.5 × 10-4 m³

Change in temperature, ΔT

=(78-5)=73°

Spillage, ΔV

=3.6 × 10-6 m³

Coefficient of volume expansion *over* that of the can

=(ΔV/V)/ΔT

=(3.6*10^(-6)/(3.5*10^(-4)))/(78-5)

=1.4090*10^(-4)

Coefficient of volume expansion of can, Cc

= 69*10^(-6)

Coefficient of volume expansion of liquid

=1.4090*10^(-4)+Cc

=2.099*10^(-4)

=2.1*10^(-4) approx.

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