When you drink cold water, your body must expend metabolic energyin order to maintain normal body temperature (37°c) by warmingup the water in your stomach. could drinking ice water, then, substitute for exercise as a way to "burn calories? " suppose youexpend 520 kilocalories during a brisk hour-long walk. how manyliters of ice water (0°c) would you have to drink in order touse up 520 kilocalories of metabolic energy? for comparison, thestomach can hold about 1 liter.

14 L

Explanation:

The ice water will have to be warmed up to 37 oC  

Quantity of heat gain the ice water = mcΔT

where m is the mass of water, c is the specific heat capacity of water

heat gain = heat loss

heat expected the body to lose = 520kcal = 520 × 4184 J ( 1 kcal is 4184 J)

520 × 4184 = 4200 × m  × 37

m = (520 × 4184) / ( 4200 × 37) where specific heat capacity of water is 4200J/KG.K

but density = m/V where V is volume in m³

m = ρ×V

replace m with the above formula

ρV = (520 × 4184) / ( 4200 × 37) and density of water is 1000kg/m³

V = (520 × 4184) / (4200 × 37 × 1000) = 0.014 m³ = 0.014 × 1000 ( 1m³ = 1000L) = 14 L

exercise will be better as it is not quite possible to drink this (14 L) volume of ice cold water.

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.

it is a negative acceleration