Answer to Question #139706 in Molecular Physics | Thermodynamics for Kgokane Sekgoka

Question

To use 100 Calories of energy, how high would an 80.0 kg person have to climb?

Answer 1

The energy, required to climb to some height, is

$E = mgh$

where g is gravitational acceleration.

We can convert given calories to metric units, namely

$E = 100 \; \textrm{cal} = 100 \cdot 4.18 \; J= 418 \;J$.

So,

$\displaystyle h = \frac{E}{mg} = \frac{418}{80 \cdot9.81} = 0.53 \, m.$

Answer: 0.53 m