Your swimming pool is square and 6.1 m on a side. It is 1.4 m deep in the morning. If the temperature changes by 19°C during the afternoon, how much does the depth of the water increase?
Use the same method I used here:
http://www.jiskha.com/display.cgi?id=1322365156
Only the pool dimensions have changed.
You need to assume a mean (average) temperature and use a mean thermal expansion coefficient for water, because it changes rapidly with temperature.
To find out how much the depth of the water increases, we need to calculate the change in the volume of the water due to the temperature change.
Let's start by calculating the initial volume of the water in the pool:
Volume = length * width * depth
Given that the pool is square with a side length of 6.1 m and a depth of 1.4 m, we can calculate the initial volume as follows:
Volume = 6.1 m * 6.1 m * 1.4 m
Next, we'll calculate the final volume of the water after the temperature change. We know that the temperature change caused the water to expand, which increases the volume. The change in volume due to thermal expansion is given by the formula:
Change in Volume = initial volume * coefficient of volume expansion * change in temperature
The coefficient of volume expansion for water is approximately 0.00021 per degree Celsius.
Given that the temperature change during the afternoon is 19°C, we can calculate the change in volume as follows:
Change in Volume = initial volume * 0.00021 * 19
Finally, to find out how much the depth of the water increases, we need to divide the change in volume by the surface area of the pool, which is the length times the width.
Change in Depth = Change in Volume / (length * width)
Substituting the values, we can calculate the change in depth as follows:
Change in Depth = (initial volume * 0.00021 * 19) / (6.1 m * 6.1 m)
Now you can plug in the values and calculate the change in depth of the water.