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Does drinking ice water burn calories?

For anyone trying to lose weight, this question is an exciting one! If you simply want to know if your body burns calories warming up the water, the answer is yes. But if you want to know if drinking a lot of ice water can help you lose weight, or keep weight off, this "yes" needs to be qualified with some calculations.

First of all, calories are case-sensitive. There are calories and then there are Calories. Calories with a big "c" are the ones used to describe the amount of energy contained in foods. A calorie with a little "c" is defined as the amount of energy it takes to raise the temperature of 1 gram of water 1 degree Celsius.

What most people think of as a Calorie is actually a kilo-calorie: It takes one Calorie to raise the temperature of 1 kilogram of water 1 degree Celsius. So when you drink a 140-Calorie can of cola, you are ingesting 140,000 calories. There is no cause for alarm, because the conversion applies across the board. When you burn 100 Calories jogging a mile, you are burning 100,000 calories.

So, considering that the definition of a calorie is based on raising the temperature of water, it is safe to say that your body burns calories when it has to raise the temperature of ice water to your body temperature. And unless your urine is coming out ice cold, your body must be raising the temperature of the water. So calories are being burned.

Let's figure out exactly what you're burning when you drink a 16-ounce (0.5 liter) glass of ice water:

  • The temperature of ice water can be estimated at zero degrees Celsius.
  • Body temperature can be estimated at 37 degrees Celsius.
  • It takes 1 calorie to raise 1 gram of water 1 degree Celsius.
  • There are 473.18 grams in 16 fluid ounces of water.
So in the case of a 16-ounce glass of ice water, your body must raise the temperature of 473.18 grams of water from zero to 37 degrees C. In doing so, your body burns 17,508 calories. But that's calories with a little "c." Your body only burns 17.5 Calories, and in the grand scheme of a 2,000-Calorie diet, that 17.5 isn't very significant.

But let's say you adhere to the "eight 8-ounce glasses of water a day" nutritional recommendation. In 64 ounces of water, there are 1,892.72 grams. So to warm up all that water in the course of a day, your body burns 70,030 calories, or 70 Calories. And over time, that 70 Calories a day adds up. So, while you definitely shouldn't depend on ice water consumption to replace exercise or a healthy diet, drinking cold water instead of warm water does, in fact, burn some extra Calories!

Source: HowStuffWorks.com
Tags: Weight Loss Health

While this information is accurate, it omits a very imporant detail. The body's primary method of controling body temperature is to regulate the amount of heat lost. The body requires a drop of at least a few degrees celsius in core temperature before it will resort to burning calories (primarily in the form of shivering) to raise temperature. Thus, the majority of heat regulation is managed through vasoconstriction/dilation and elimation of waste produces. That said, let us consider our example of drinking a 16oz. glass of water.

As calcualted above using the formula E=mcΔT, it was accurately calculated that to raise 16oz. of ice water to body temperature would require absorption of approximately 17.5 Calories... However, assuming a 70kg body mass (approximately 154 pounds), and using the same forumula used earlier, the effect on body temperature is shown below.

Specific heat of the body (c)= .829 Cal/(kgºC)
Energy lost (E) = 17.5 Cal
Mass (m)= 70kg
Change in body temperature (ΔT) = ?

E = mcΔT

can be rewritten to solve for ΔT

ΔT = E/mc

Substituting values for variables we get

ΔT = 17.5 Cal/(70 kg * .829 cal/(kgºC))


ΔT ≈ 0.3ºC ≈ .54ºF

Lowering the body temperature by .3ºC (.54ºF) is not sufficient to induce a shivering response, and thus, the temperature difference will be managed by a more passive vasoconstriction resulting in a temporary reduction in the amount of heat lost until body temperature returns to its optimal state. However, because the temperature change will actually be in a gradient radiating outward from the stomach, it is possible that there might be a slight consumption of calories to account for the sharper change temperature in immediate proximity to the stomach, but this will be nowhere near the 17.5 Cal mentioned above. However, without any experimental data, this is all just conjecture based on theories which are oversimplified and often do not account for variaions that occur in the real world.

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