Question

Consider a corn yield of 7,500 kg/hectare (equivalent to 120 bushels per acre). If 25 kg (one bushel) of corn consumed about 20m3 of water during the growing season, what is the ratio of the weight of corn to the weight of water consumed

Answer 1

Answer:

$\frac{W_{1}}{W_{2}} = 0.5$

Explanation:

The ratio of the weight of corn to the weight of water consumed will be the same. Either we compare the total weights for the total amount of corn or we compare only a small proportion of it. So, we consider the given 25 kg of corn. So, the mass of corn is given:

m₁ = mass of corn = 25 kg

Now, for mass of water, we will use its density:

ρ₂ = m₂/V₂

m₂ = ρ₂V₂

where,

m₂ = mass of water = ?

V₂ = Volume of water consumed by 25 kg corn = 20 m³

ρ₂ = density of water = 1000 kg/m³

Therefore,

m₂ = (1000 kg/m³)/(20 m³)

m₂ = 50 kg

Now, we calculate the ration:

$\frac{Weight of Corn}{Weight of Water} = \frac{W_{1}}{W_{2}} = \frac{m_{1}g}{m_{2}g}$

$\frac{W_{1}}{W_{2}}=\frac{m_{1}}{m_{2}} = \frac{25 kg}{50 kg}$

$\frac{W_{1}}{W_{2}} = 0.5$