About how many calories will I use if I drink 1 liter of ice water?
103149
If you drink 1 liter of ice water, your body will use approximately 35 calories to raise it up to your internal body temperature.
673582
This information should enable you to solve the problem. \n\nLatent Heat of Freezing or Solidification \n\nThe amount of heat required per unit mass of a substance at freezing point (constant) and under 1 Atmosphere pressure to convert it from the liquid to the solid state is called Latent Heat of Freezing.\n\n \ne.g. At 00C & 1 Atmosphere pressure 80 K-calorie of heat supplied to 1 Kg of ice melts it into water and 80 K-Calorie of heat taken away from 1 Kg of water converts it into ice. \n\n&#92; L = 80 Kcal / Kg = Latent heat of fusion of ice\n\n\n    Latent heat of freezing of water \n\nso if 80 Kcal/Kg= latent heat of fusion of ice\nthen you can cross out the K and you have 80 cal/gram is the heat of fusion of water. but you hve 25 grams of water to freeze, not just 1 gram. so what comes next. what do you have to multiply by 25?\n80 right? \nthats right. see it wasnt so hard was it?\nso the answer is 2000calories or 2 Kcal.
211914
It depends on the mass. The latent heat of fusion of water (the energy required to freeze it) is 333.7 kJ/kg. That means to freeze one kilogram of water, it requires removing 333.7 kilojoules of energy.\n\nThis assumes that the water is already at freezing temperature. If not, energy must first be removed to get it at freezing temp.
22647
There's a formula in this article:\nhttp://www.naturodoc.com/library/nutrition/water.htm
725897
contrary to the other answer, DO drink COLD water. your metabolism will need to work harder to heat it up, causeing you to burn calories. also, run for 30 minutes in the morning and 30 at night. it works!
175212
Only water weight, which means your going to gain it right back as Soon as you drink anything...so no. Sorry. Would be nice though wouldn't it? \nInteresting note that kinda contradicts the whole sauna thing: Drinking cold water burns slightly more calories then drinking warm or room temperture water because your body has to use energy to heat it up and maitain your overall body temperature. That's key, your body has to use more energy then it takes in to loose weight. You aren't using any energy in a suana. Just sweating off water weight.
755464
you need to find how many calories the exercise burn, find your resting metabolic rate, subtract 500 calories from that, 500 calorie reduction cause weight loss of 1 lb, but be sure to take in min of 1200 calories per day for healthy food and protein,  and half you body weight in oz. per day of water, protein is your energy source when exercising if you do not have proper protein intake you body will burn the muscle convert to fat and use that for energy, if you take in protein your body will burn the fat to help process the protein thus leaving you lean and not fluffy.
356284
Depends on weight, height and all that other stuff, but here is a calculator for you:\n\nhttp://www.freedieting.com/tools/calorie_calculator.htm
978927
You have to be careful about what you mean by calorie, as there are two kinds.  The standard calorie (the energy it takes to raise 1 cc of water 1ºC) is 4.185 joules.  However the calorie used to depict the energy content of food is the "large Calorie", spelled with a capital "C",  and is equal to one kilocalorie and therefore 4.185 kilojoules.
507893
1 calorie heats 1g of water by 1 deg C.\n630cal heats 630g of water 1 deg C\n630cal heats (630/15)g water by 15deg
528628
maybe 1000-1200 calories a day, no carbs, lots of water.\n\nrun at least 4-5 miles per day.
1177074
A calorie is the quantity of thermal energy required to raise one gram of water 1°C at 15°C.\n\nCarbohydrates have 4 calories per gram. Proteins have 4 calories per gram. Alcohol has 7 calories per gram. Fat has 9 calories per gram. Knowing this, you just add up the calories to get the total amount in an item of food/drink.
1069522
Drink water but put that Crystal light in it and make a batch, YUM... you think you're drinking kool-aid and NO calories
1119466
I donno ;=)\n\nMay be start with a heat balance equation like\n\nQfinal=Qwater-start  - Qice-start – Qlatent\n\nIn other words the final heat of the system is equal to the original amount of heat minus the heat required to bring ice to the melting point and minus again the amount of heat needed to melt the ice\n\nOr \nQfinal=(Specific-heat of water) (temp in K)(mass-final)\n\nmass-final= mass of water + mass of ice= m1 + m2\n\nNote -273.16 °C =0 degrees on Celsius scale = 273.16 °K (273 I hope is good enough for you?)\n\nQwater-start  =( 4190 J/(kg*K) (t-start K degrees )(m1)( 4190 J/(kg*K)=A(m1)  A= 4190 (76.2 + 273)\n\nQice-start= 2100 J/(kg*K)( - 23.6+273)(m2)= B(m2)\n\n\nQlatent= 334 kJ/kg(m2) =C(m2)\n\nWe have \n\nD(m1+m2)=A(m1)+B(m2)+C(m2)  you know that m1=.280kg  then you figure out m2? \nAnd \nA= 4190 (76.2 + 273)\nB=2100 (- 23.6+273)\nC=334,000\n\nActually Specific heat of ice = 2060 J/(kg*K) not 2100 J/(kg*K)\nAnd Specific heat of water  =4,183  J/(kg*K)
739180
Warm water is absorbed easier into the body.  Cold water takes longer to absorb, therefore burning more calories.
512779
Well... break it down to formulas. 1 calorie is the energy needed to heat 1 gram of water by 1 degree celcius.\n\nSo to heat 1000 grams of water (1 liter) by 1 degree, it takes 1000 calories. Now double the water to 2 liters and it will take 2000 calories.\n\nIf you are adding heat at a rate of 1000 calories per minute, then 1 liter of water heats at 1 degree celcius per minute and 2 liters heat at a rate of 1/2 degree celcius per minute. So yes. It is linear.\n\nHOWEVER (there is always a however) there are many factors that go into this in a lab setup. Namely, a larger volume of water will have more surface area and disipate heat to the suroundings faster. Also, a larger pan (wider) will heat faster than a taller pan. Etc, etc, etc.
1110819
Not many about 50 calories. I know this as I had my polar watch on & was disappointed. But then again it would depend upon how long and fitness levels on each & your heart rates.