Question

An athlete is 18% efficient performing a specific exercise. They eat an apple,
which delivers 490 kcal to their body. For how many minutes must they exercise,
generating mechanical energy at a rate of 200 W, to “burn off” the calories contained in
the apple?

Answer 1

Energy conversion efficienty of athlete:

$\eta=\frac{P_{mech}}{P_{internal}} =\frac{P_{mech}}{Q_{apple}/t_{exercise}}=\frac{P_{mech}\times t_{exercise}}{Q_{apple}}$

taking into account that 1 kcal = 4186.8 J :

$t=\frac{\eta \times Q_{apple}}{P_{mech}}=\frac{0.18\times490\times4186.8}{200}=1846sec=30.77min$