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.
To determine how much the depth of the water increases, we first need to find the coefficient of thermal expansion for water. The coefficient of thermal expansion for water is approximately 0.00021 per degree Celsius.
Given that the temperature changes by 14°C during the afternoon, we can calculate the change in volume of the water using the formula:
Change in volume = Coefficient of Thermal Expansion * Original Volume * Change in Temperature
To find the original volume of the water, we can use the formula:
Original Volume = Length * Width * Depth
Given that the swimming pool is square and 5.7 m on a side, the length and width are both 5.7 m. The initial depth of the water is 1.5 m.
Plugging in the values, we have:
Original Volume = 5.7 m * 5.7 m * 1.5 m
Change in volume = 0.00021 (1/°C) * (5.7 m * 5.7 m * 1.5 m) * 14 °C
Simplifying the calculations, we have:
Original Volume = 48.915 m³
Change in volume = 0.00021 * 48.915 * 14
Change in volume = 0.14565 m³
Therefore, the depth of the water increases by approximately 0.146 meters (or 14.65 centimeters) during the afternoon.