Question

1. Suppose during volleyball practice, you lost 2.0 lbs of water due to sweating. If all of this

water evaporated, how much energy did the water absorb from your body? Express your

answer in kJ. (2.2 lbs = 1.0 kg)

Answer 1

Answer:

$E=2052.8 kJ$

Explanation:

The energy absorbed by the water is the energy it requires to evaporate. So:

$Heat of .vap= 40.65 kJ/mol$

The moles of water:

$n_w=m_w *\frac{1000g}{2.2lbs}*\frac{1}{M}$

M is the water molecular weight

$n_w=2 lbs *\frac{1000g}{2.2lbs}*\frac{1}{18g/mol}$

$n_w=50.5mol$

The energy absorbed:

$E=n_w*Heat of vap.=50.5mol *40.65 kJ/mol$

$E=2052.8 kJ$