Question

3.00 m^3 of water is at 20.0°C.

Answer 1

Answer:

$\triangle V = 0.02484m^3$

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 $0.02484m^3$