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The food calorie, equal to 4,186 J, is a measure of how much energy is released when food is...

Question:

The food calorie, equal to 4,186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.

Part A

If a 61-kg hiker eats one of these bars, how high a mountain he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential erengy?

Part B

If, as is typical, only 25% of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)

Energy:

Energy is the potential of an object to perform work. It comes in many different forms. It can be inside the food we eat as chemical energy. It can be kinetic energy as we mobilize ourselves using the chemical energy from our food. It can be stored again, but now, as potential energy, once we reach a higher location in reference to our previous one. As human beings attempt to accomplish more and more work, we started devising ways to harness the energy we find in the physical world. This is how we now have the convenience of electricity, of refrigeration, cooling and heating, and many others that have bettered our lives.

Answer and Explanation:

Given:

{eq}E_b = 160 \ calories \times \dfrac {4186 \ J}{1 calorie} = 669760 J \\ m = 61 \ kg {/eq}

Find:

Part A

If a 61-kg hiker eats one of these bars, how high a mountain he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential erengy?

To expend all the calories, we must equate

{eq}E = PE = mgh {/eq}

Now, using this, we can get the height required

{eq}h = \dfrac {E}{mg} = \dfrac {669760 \ J}{(61 \ kg)(9.81 \ \dfrac {m}{s^2}} = 1119.23 \ m {/eq}

Part B

If, as is typical, only 25% of the food calories go into mechanical energy, what would be the answer to part (a)?

Using the same formula but with 25% of the energy to consume through mechanical work

{eq}h = 25 \% \dfrac {E}{mg} = 25 \% \dfrac {669760 \ J}{(61 \ kg)(9.81 \ \dfrac {m}{s^2}} = 279.81 \ m {/eq}


Learn more about this topic:

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Energy Conservation and Energy Efficiency: Examples and Differences

from Geography 101: Human & Cultural Geography

Chapter 13 / Lesson 9
100K

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