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?

Respuesta :

Answer:

0.625 (62.5 %)

Explanation:

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

[tex]\eta = \frac{E_{useful}}{E_{in}}[/tex]

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,

[tex]\eta = \frac{750 J}{1200 J}=0.625[/tex]

Which means an approximate efficiency of 62.5 %.