Question

Please explain too

A student evaluates a weight loss program by calculating the number of times she would need to climb a 15.0 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 21.2% 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 96.0 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 of fat can release an energy of 3.77 * 10⁷ J
0.450 kg of fat can release an energy of 0.45 * 3.77 * 10⁷ J
= 1.697 * 10⁷ J

21.2% of total energy released is converted into mechanical energy
Mechanical energy produced by 1.697 * 10⁷ J of energy, E = 21.2% * 1.697 * 10⁷ J
= 3.60 * 10⁶ J

B)
Energy equivalent of climbing 15 m, Q = m * g * h
Where m is the mass of the student and h is the height.
Q = 96 * 9.81 * 15
= 1.41 * 10⁴ J

Number of times needed to lose the 0.45 kg of fat, n = E/Q
= (3.60 * 10⁶ J) / (1.41 * 10⁴ J)
= 255 times