Answer to Question #240854 in Molecular Physics | Thermodynamics for Kelani

Question

A student evaluates a weight loss program by calculating the number of times he would need to climb a 13.0 m high flight of steps in order to lose one pound (0.45 kg) of fat. Metabolizing 1.00 kg of fat can release 3.77 ✕ 10⁷ J of chemical energy and the body can convert about 22.0% of this into mechanical energy (the rest goes into internal energy.)

(a) How much mechanical energy (in J) can the body produce from 0.450 kg of fat?

__ J

(b) How many trips up the flight of steps are required for the 92.0 kg student to lose 0.450 kg of fat? Ignore the relatively small amount of energy required to return down the stairs.

 __trips

Answer 1

(a) 1 kg produces 3.77 ✕ 10⁷ J of chemical energy, and only 22% of the 3.77 ✕ 10⁷ J are mechanical energy, therefore

$$E_{M1}=3.77·10^7·0.22=8294000\text{ J}.$$

Therefore, 0.45 kg are produce

$$E_{M}=0.45E_M=3732300\text{ J}.$$

(b) The students wants to burn fat doing mechanical work:

$$W=nmgh,$$

where n - the number of times he climbed the stairs.

Equate this work to mechanical energy produced by 450 g of burned fat:

$$W=E_M,\\\space\\ nmgh=E_M,\\\space\\ n=\frac{E_M}{mgh}=\frac{3732300}{92·9.8·13}=319.$$

Quite many times. This is equivalent to 319·13=4.15 km of walking upstairs or 8.3 km of walking in total.