Question

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)

Answer 1

9m^3

Step-by-step 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