Question

A sparrow eats 1200 j of seeds. of this, 425 j is eliminated as feces and 750 j are used in cellular respiration. what is the approximate production efficiency of this animal?

Answer 1

0.625 (62.5 %)

Step-by-step explanation:

The production efficiency is equal to the rate between the energy converted into useful energy and the energy in input:

`\eta = (E_(useful))/(E_(in))`

where in this case:

- the useful energy is 750 J (the energy used in cellular respiration)

- the energy in input is 1200 J (the energy taken by eating the seeds)

Substituting,

`\eta = (750 J)/(1200 J)=0.625`

Which means an approximate efficiency of 62.5 %.