3.00 m^3 of water is at 20.0°C. If you raise its temperature to
60.0°C, by how much will its
volume expand?
Water
ß = 207•10-6C-1
(Unit = m^3)

9m^3

Explanation:

Given data

volume  v1=  3m^3

volume  v2=  ???

Temperature T1= 20.0°C.

Temperature T2= 60.0°C.

Applying the relation for temperature and volume

V1/T1= V2/T2

substitute

3/20= V2/60

3*60= V2*20

180= 20*V2

180/20= V2

V2= 9m^3

Hence the final volume is 9m^3

it is the second option.

it is called kinetic energy