Question

3.00 m^3 of water is at 20.0°C.If you raise its temperature to60.0°C, by how much will itsvolume expand?WaterB = 207•10-6 0-1(Unit = m^3)

Answer 1

Given,

Initial volume of the water, V₁=3.00 m³

The initial temperature of the water, T₁=20.0 °C=293.15 K

The final temperature of the water, T₂=60 °C=333.15 K

From Charle's law, we have,

`\frac{V_1}{T_1_{}}=(V_2)/(T_2)`

On rearranging the above equation,

`V_2=(V_1T_2)/(T_1)`

On substituting the known values in the above equation,

`V_2=(3.00*333.15)/(293.15)=3.41m^3`

Therefore the change in the volume is,

`\Delta V=V_2-V_1`

i.e.,

`\Delta V=3.41-3.00=0.41m^3`

Therefore, the volume of the water will expand by 0.41 m³