Question

When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature (37° C) by warming up the water in your stomach. Could drinking ice water, then, substitute for exercise as a way to "burn calories?" Suppose you expend 450 kilocalories during a brisk hour-long walk. How many liters of ice water (0° C) would you have to drink in order to use up 450 kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter.

Answer 1

Answer:

$V=12.1622\ L$

Explanation:

Given:

• Initial temperature of water, $T_i=0^{\circ}C$

• final temperature of water, $T_f=37^{\circ}C$

• energy spent in one hour of walk, $Q=450\ kcal=450\times4186=1883700\ J$

• volumetric capacity of stomach, $V=1\ L$

Now, let m be the mass of water at zero degree Celsius to be drank to spend 450 kilo-calories of energy.

We know:

$Q=m.c.\Delta T$ .....................................(1)

where:

m = mass of water

Q = heat energy

$\Delta T=$ temperature difference

$c=4186\ J.kg^{-1}.K^{-1}$ = specific heat capacity of water

Putting values in the eq. (1):

$1883700=m\times 4186\times 37$

$m=12.1622\ kg$

Since water has a density of 1 kilogram per liter, therefore the volume of water will be:

$V=12.1622\ L$