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.