Physics -- Temp and heat
posted by Anon14 on .
Your swimming pool is square and 5.7 m on a side. It is 1.5 m deep in the morning. If the temperature changes by 14°C during the afternoon, how much does the depth of the water increase?
5.7^2 * 1.5 m^3 is the volume in the morning, Vo.
In the afternoon, the volume increases by a factor V/Vo = K*delta T
where K is the bulk thermal expansion coefficient of water. You will need to look that up. DeltaT = 14 C
You will find that K depends upon temperature. To treat is as a constant, you need some idea of the average termperature. At about 20 C average temperature, K = 2*10^-4
The pool depth will increase by the same ratio V/Vo, since the horizontal area will not change. Use that fact to deduce the depth change.
I get about 4 millimeters depth increase. Evaporation has been neglected.