Question

b) How many days of a diet of 2200 large calories are equivalent to the gravitational energy change from sea level to the top of Mount Everest, 8848 m above sea level? Assume your mass is 72 kg. (The body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)

Answer 1

To find the number of diet days equivalent to the gravitational energy change to the summit of Mount Everest, the gravitational potential energy is calculated and then converted to Calories. It would take roughly 6.73 days of a 2200 Calorie diet to equal the energy required for this ascent.

Step-by-step explanation:

The student has asked how many days of a diet of 2200 large calories are equivalent to the gravitational energy change from sea level to the top of Mount Everest. To solve this, first we need to calculate the gravitational potential energy change. The gravitational potential energy (GPE) is given by the formula GPE = mgh, where m is mass in kilograms, g is the acceleration due to gravity (9.8 m/s2), and h is height in meters.

For a mass of 72 kg and height of 8848 m, the GPE change would be GPE = 72 kg * 9.8 m/s2 * 8848 m. This results in a GPE change of approximately 61926336 Joules.

Now, knowing that 1 large calorie (Cal) is equivalent to 4184 Joules, we can convert the GPE change into Calories by dividing 61926336 J by 4184 J/Cal, which gives approximately 14802 Calories.

Finally, assuming a diet of 2200 large Calories per day, we divide 14802 Calories by 2200 Calories/day to find the number of days. This results in approximately 6.73 days. Therefore, it would take roughly 6.73 days of a 2200 Calorie diet to equal the energy required to ascend to the summit of Mount Everest.

Answer 2

No. of days, n = 0.68 days

Solution:

As per the question:

Energy of one day diet,
`E_(d) = 2200\ large\ calories`

Height above sea level, H = 8848 m

Mass, M = 72 kg

Now,

1 large calorie =
`4.2* 10^(3) J`

Therefore,

`E_(d) = 2200* 4.2* 10^(3) = 9.24* 10^(6)J`

Now,

The potential energy required for jumping of the Mt. Everest at a height of 8848 m:

`PE_(g) = MgH = 72* 9.8* 8848 = 6.24* 10^(6) J`

No. of days can be given as:

`n = (PE_(g))/(E_(d))`

`n = (6.24* 10^(6))/(9.24* 10^(6)) = 0.68\ days`