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 286 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 286 kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter.

Answer 1

Answer:

7.72 Liters

Explanation:

normal body temperature = T_body =37° C

temperature of ice water = T_ice =0°c

specfic heat of water = c_{water} =4186J/kg.°C

if the person drink 1 liter of cold water mass of water is = m = 1.0kg

heat lost by body is Qwater =mc_{water} ΔT

                                           = mc{water} ( T_ice - T_body)

                                             = 1.0×4186× (0 -37)

                                             = -154.882 ×10^3 J

here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories

            = 286×4186J

so number of liters of ice water have to drink is

n×Q_{water} =Q_{walk}                                                                        n= Q_{walk}/ Q_{water}

= 286×4186J/154.882×10^3 J

 = 7.72 Liters

Answer 2

Answer:

$V=7.73\ L$

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, $286\ kilocal=(286\times 4186)\ J$

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

Now, let m be the mass of water at zero degree Celsius to be drank to spend 286 kilo-calories of energy.

$\therefore Q=m.c_w.\Delta T$ .....................................(1)

where:

m = mass of water

Q = heat energy

$c_w=4186\ J\ (specific\ heat\ of\ water)$

$\Delta T$= temperature difference

Putting values in the eq. (1):

$286\times 4186=m\times 4186\times 37$

$m=7.73\ kg$

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

$V=7.73\ L$