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
B = 207•10-6 0-1
(Unit = m^3)

Answer 1

`\triangle V = 0.02484m^3`

Step-by-step explanation:

Given

`V_1 = 3.00m^3` --- initial volume

`T_1 = 20.0^oC` --- initial temperature

`T_2 = 60.0^oC` --- final temperature

`\gamma = 207*10^{-6` --- coefficient of thermal expansion:

Required

The change in volume

To do this, we make use of cubic expansivity formula

`\triangle V = \gamma * V_2 * (T_2 - T_1)`

So, we have:

`\triangle V = 207 * 10^(-6) * 3.00 * (60.0 - 20.0)`

`\triangle V = 207 * 10^(-6) * 3.00 * 40.0`

`\triangle V = 0.02484m^3`

The volume will expand by
`0.02484m^3`