Question

When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature (37.0°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 510. 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 510. kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter. Number Units

Answer 1

let m is the ice water reuired.

given

Q = 510 kcal

= 510*10^3*4.186 J

we know,

specific heat of water, C = 4186 J/(kg C)

T1 = 0 C
T2 = 37 C

now use, Q = m*C*(T2 - T1)

==> m = Q/(C*(T2 - T1)

= 510*10^3*4.186/(4186*(37 - 0))

= 13.8 kg

= 13.8 L (since 1 L is equivalent to 1 kg of water)