Question 17.2: Everyone has heard about the food chain, but few realize how...
Everyone has heard about the food chain, but few realize how inefficient it is in nature. The energy conversion efficiency from sunlight to plant growth is only about 1.00\%, the energy conversion efficiency of the plants eaten by grazing herbivores is about 20.0\%, and the energy conversion efficiency of the carnivores who hunt and eat the herbivores is only about 5.00\%. So the overall energy conversion efficiency from sunlight to carnivore is about (0.0100)(0.200)(0.0500) = 1.00 × 10^{−4} = 0.0100\%. If the average daily solar energy reaching the surface of the Earth is 15.3 \text{MJ/d. m}^2 , then how much land is required to grow the plants needed to feed the herbivores eaten by a large carnivore that requires 10.0 \text{MJ/d} to stay alive?
Since our hunting carnivore requires 10.0 \text{MJ} of food per day, at a 5.00\% food energy conversion rate, it must consume
\frac{10.0 \text{MJ/d}}{0.0500} = 200. \text{MJ/d}
of herbivore meat. The food energy conversion rate of the grazing herbivores is 20.0\%, so they must consume
\frac{200. \text{MJ/d}}{0.200} = 1000 \text{MJ/d}
in plant food. At a 1.0\% energy conversion rate, the plants consumed by the herbivore require
\frac{1000 \text{MJ/d}}{0.0100} = 1.00 × 10^5 \text{MJ/d}
of solar energy. Since the average solar energy intensity on the surface of the Earth is 15.3 \text{MJ/d.m}^2 , 100,000 \text{MJ/d} of solar energy require an area of
\frac{100,000 \text{MJ/d}}{15.3 \text{MJ/d.m}^2} = 16540 \text{m}^2
and since 1 \text{acre} = 4047 \text{m}^2, then
6540 \text{m}^2 (\frac{1 \text{acre}}{4047 \text{m}^2}) = 1.62 \text{acres}
of plant food is required to supply the food chain energy required to meet the 10 \text{MJ/d} needs on our carnivore.