Answer to Question #96627 in General Chemistry for LK

Question

A 41-kg person drinks 570. g of milk, which has a "caloric" value of approximately 3.0 kJ/g. If only 16% of the energy in milk is converted to mechanical work, how high (in meters) can the person climb based on this energy intake?[Hint: The work done in ascending is given by mgh, where m is the mass (in kilograms), g the gravitational acceleration (9.8 m/s2), and h the height (in meters.)]

Answer 1

Q96627

Solution:

Mass of the person$=m=41kg.$

Amount of milk taken $=570g$

Calorific value of milk $=3kJ/g$

Thus, $Energy consumed =amount of milk*calorific value$

$=(570g)*(3kJ/g)$

$=1710kJ.$

Given, only of 16% of the consumed energy is converted to Mechanical energy.

Thus, Mechanical energy$=(16/100)*1710kJ$

$=273.6kJ$

$=mgh$

where, $g=9.8m/s^2$

Thus, $h=273.6*10^3/(mg)$

$=(273.6*10^3)/(41*9.8)$

$=680.936m$ (Answer)

Thus, the person can climb upto a height of $680.936m$ based on this energy intake.