Your swimming pool is square and 6.0 m on a side. It is 3.7 m deep in the morning. If the temperature changes by 37°C during the afternoon, how much does the depth of the water increase?
Question
Your swimming pool is square and 6.0 m on a side. It is 3.7 m deep in the morning. If the temperature changes by 37°C during the afternoon, how much does the depth of the water increase?
Solution
To solve this problem, we need to use the formula for the volume expansion of liquids. The formula is:
ΔV = βV0ΔT
where:
- ΔV is the change in volume,
- β is the coefficient of volume expansion of the liquid (in this case, water),
- V0 is the initial volume, and
- ΔT is the change in temperature.
The coefficient of volume expansion of water is approximately 0.000214 per degree Celsius.
First, we calculate the initial volume of the water in the pool:
V0 = side^2 * depth = 6.0 m * 6.0 m * 3.7 m = 133.2 m^3
Then, we calculate the change in volume:
ΔV = βV0ΔT = 0.000214 1/°C * 133.2 m^3 * 37 °C = 1.06 m^3
Since the pool is square, the increase in depth (Δh) is equal to the change in volume divided by the area of the pool:
Δh = ΔV / (side^2) = 1.06 m^3 / (6.0 m * 6.0 m) = 0.03 m
So, the depth of the water increases by 0.03 m (or 3 cm) when the temperature changes by 37°C during the afternoon.
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