   Chapter 5, Problem 74AP

Chapter
Section
Textbook Problem

A 50.0-kg student evaluates a weight loss program by calculating the number of times she would need to climb a 12.0-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 20.0% of this into mechanical energy. (The rest goes into internal energy.) (a) How much mechanical energy can the body produce from 0.450 kg of fat? (b) How many trips up the flight of steps are required for the student to lose 0.450 kg of fat? Ignore the relatively small amount of energy required to return down the stairs.

(a)

To determine
The mechanical energy that can be produced from 0.450kg of fat.

Explanation

Given Info:

1.0kg of fat can release 3.77×107J of chemical energy

The body can convert 20% of the chemical energy to mechanical energy

Formula to calculate the mechanical energy produced is,

Emec=(0.20)Echm       (I)

• Echm is the chemical energy available from 0.450kg of fat

Since,

Echm=mf(3.77×107J1.0kg)       (II)

• mf is the mass of fat

On substituting equation (II) in (I),

Emec=(0.20)(mf)(3

(b)

To determine
The number of trips up the flight of steps are necessary for the student to lose 0.450kg of fat.

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