The tutor’s casual observation leads to a discussion about evaporation.

Last Saturday, my children and I had a water fight in the back yard. We filled two buckets (one for each side) from which to replenish our water guns.

Looking outside today, I noticed the water in one of the buckets has lost about 2cm depth. Its diameter is 24cm (radius=12cm). What about a quick calculation of the rate the water is evaporating?

First, we find the volume that has evaporated. It’s a cylindrical shaped column of height 2cm:

V = Πr²h = Π(12)²(2)= 905cm³ = 905mL

The density of water is 1g/mL. Therefore,

905mL=905g

It’s been about three hours short of four days since the water fight. Let’s call it

4×24-3=93h

The evaporation rate per hour is

905g/93h=9.73g/h or 9.73mL/h

For any circular water surface in this context, the general formula for the evaporation rate is

9.73g/(hourx12²) x R²=0.0676g/hour x R², R=surface radius in cm

BTW: the average temp, day and night, has been around 16°C since the bucket was filled:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.