3 years ago

3.00 m^3 of water is at 20.0°C.

Physics

1 answer:

krok68 [10]3 years ago

6 0

Answer:

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

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