Question

A heavy rainstorm dumps 1.0 cm of rain on a city 7 km wide and 4 km long in a 2 h period.

How many metric tons ( 1 metric ton = 103kg) of water fell on the city? (1 cm3 of water has a mass of 1 g = 10−3kg.)

How many gallons of water was this?

Answer 1

Volume of the water dumped will be

Volume = length*width*height

Length = 7 km = 7*10^5 cm

width = 4 km = 4*10^5 cm

height = 1.0 cm

So,

Volume = (7*10^5 cm)*(4*10^5 cm)*(1.0 cm)

Volume = 28*10^10 cm^3 = 2.8*10^11 cm^3

Mass of this water will be

mass = density*Volume

density of water = 1 gm/cm^3

mass = (1 gm/cm^3)*(2.8*10^11 cm^3)

mass = 2.8*10^11 gm = 2.8*10^8 kg

1 metric ton = 10^3 kg

mass = 2.8*10^5*10^3 kg = 2.8*10^5 metric ton

Part B.

we know that

1 cm^3 = 1 mL

1 L = 0.264172 gallon

So,

Volume of water = 2.8*10^11 cm^3 = 2.8*10^11 mL = 2.8*10^8 L

Volume of water = (2.8*10^8 L)*(0.264172 gallons/1 L) = 7.3968*10^7 gallons

Volume of water = 7.4*10^7 gallons

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