Answer:
a) The energy efficiency of the out-of-condition professor is 9.082 %.
b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
Explanation:
a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:
[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)
Where:
[tex]W[/tex] - Useful work, in joules.
[tex]E[/tex] - Food energy, in joules.
If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:
[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]
[tex]\eta = 9.082\,\%[/tex]
The energy efficiency of the out-of-condition professor is 9.082 %.
b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:
[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]
[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]
[tex]E = 7.60\times 10^{5}\,J[/tex]
[tex]E = 181.644\,kcal[/tex]
The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
