You are attempting to water some flowers in a garden and only have a leaky watering can to do so. After consulting your l°Cal agronomist, you know that the flowers need 2.1 kg of water to flourish. The watering can holds, when filled, 5.0 liters of water. The water leaks from the can at a rate of 0.12 kg/s. It takes 30 seconds to walk from the spigot where you fill the watering can to the brim to the flowers. Consider any water that leaks from the can while you are watering the flowers falls onto the flowers. How many trips between the spigot and the flower bed will you need to make in order to give the flowers the proper amount of water?

Question

Given: V = 5 l = 0.005 m3; mreq = 2.1 kg; m?out= 0.12 kg/s; t = 30 s

What will be an ideal response?

Answer 1

p=1000 kg/m3, so the mass of water in the can is mcan = pV = 5 kg
The mass of water that leaks out of the can in one trip from the spigot to the flowers is
mloss = m?outt = (0.12 kg/s) (30 s) = 3.6 kg
Therefore, the amount of water remaining in the can when one reaches the flowers is
mleft = mcan – mloss = 5 kg – 3.6 kg = 1.4 kg
As you need 2.1 kg of water for the flowers, you will need to make 2.1/1.4 = 1.5 trips, but as you can not make a half of a trip, you will need to make 2 trips to water the flowers.