A typical person breathes in and exhales 8 liters of air per minute. On a cold winter day, the air comes in cold and dry; it leaves warm and moist - that's why you can "see your breath" on a cold day, as the exhaled moisture condenses.
If you breathe in 1.0 liter of cold, dry air, you lose about 0.040 g of water to evaporation. This evaporation of moisture from the lungs constitutes an energy loss. Breathing at a steady rate on a winter day entails a certain rate of energy loss, a certain power, which we can express in watts. What is this power (in W)?
If you breathe in 1.0 liter of cold, dry air, you lose about 0.040 g of water to evaporation. This evaporation of moisture from the lungs constitutes an energy loss. Breathing at a steady rate on a winter day entails a certain rate of energy loss, a certain power, which we can express in watts. What is this power (in W)?
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8 liters / minute x 0.040 grams / liter gives 0.32 grams per minute or converting to seconds gives 2 / 375 grams per second. Just multiply that number by joules per gram and you have watts and you're already done.
So you're missing that piece of info, the joules (energy) per gram of water. It's either in the problem statement or you can likely assume some data from an example in your book.
For example you could model it as losing the energy it took to heat the water in the first place. The body runs at about 37 degrees Celsius, and to increase 1 gram of water 1 degree takes 1 calorie or about 4.7 joules. From freezing to body temp therefore would take 173.9 joules per gram. Joules per gram times the 2/375 grams per second gives you Joules per second (which is watts which is power).
Of course that's assuming all your water intake was done with ice :p Not likely, but you see the idea. Relate Joules (energy) to grams of water, multiply by 2/375, and you're done.
So you're missing that piece of info, the joules (energy) per gram of water. It's either in the problem statement or you can likely assume some data from an example in your book.
For example you could model it as losing the energy it took to heat the water in the first place. The body runs at about 37 degrees Celsius, and to increase 1 gram of water 1 degree takes 1 calorie or about 4.7 joules. From freezing to body temp therefore would take 173.9 joules per gram. Joules per gram times the 2/375 grams per second gives you Joules per second (which is watts which is power).
Of course that's assuming all your water intake was done with ice :p Not likely, but you see the idea. Relate Joules (energy) to grams of water, multiply by 2/375, and you're done.
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The energy loss has nothing to do with the amount of water being evaporated, but rather the increase in temperature of the air out vs air in.
Assuming all the air in reaches body temperature (37°C), the heat exchange H = ρ*(dv/dt)*Cp(37 - Ta) = 1.29*(.008*60)*1006*(Ta - 37) J/sec or watts
So, to answer your question, you MUST know Ta
Assuming all the air in reaches body temperature (37°C), the heat exchange H = ρ*(dv/dt)*Cp(37 - Ta) = 1.29*(.008*60)*1006*(Ta - 37) J/sec or watts
So, to answer your question, you MUST know Ta