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.

multiply 2kg by latent heat of vap

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it