If a person breathes 10 liters per minute of air at 68 degreesF and 50% relative

Question

If a person breathes 10 liters per minute of air at 68 degreesF and 50% relative humidity, how much water per day must theinternal membranes supply to saturate the air at 98.6 degrees F? (Assume all the moisture is exhaled) If each gram of water extracts580 calories as it is vaporized, how much daily heat loss inkilocalories (food calories) does this represent? (Saturation vaporpressure at 20 degrees C is 17.3 g/m cubed and 37 degrees C is 44.0g/m cubed (grams per cubic meter). Answer Water per day: _________grams Energy per day lost:   _______ If a person breathes 10 liters per minute of air at 68 degreesF and 50% relative humidity, how much water per day must theinternal membranes supply to saturate the air at 98.6 degrees F? (Assume all the moisture is exhaled) If each gram of water extracts580 calories as it is vaporized, how much daily heat loss inkilocalories (food calories) does this represent? (Saturation vaporpressure at 20 degrees C is 17.3 g/m cubed and 37 degrees C is 44.0g/m cubed (grams per cubic meter). Answer Water per day: _________grams Energy per day lost:   _______

Explanation / Answer

l0 liters / min = 600 l / hr = 14,400 l /day 68 F = 20 C & 98.6 F = 37 C Total volume exchanged per day is V = 14.400L/d = 14.4m³/d Inhaled air is 50% saturated at 20°C, that means water contentis c_in = 0.5 · 17.3g/m³ = 8.65g/m³ So the net flow of water into the body is m_in = c_in · V = 14.4m³/d · 8.65g/m³ =124.56g/d Exhaled air is 100% saturated at 37°C, that means water contentis c_ex = 44.0g/m³ So the net flow of water into the body is m_ex = c_ex · V = 14.4m³/d · 44.0g/m³ =633.6g/d So the mass of water evaporated at the internal membranes is: ?m = m_ex - m_in = 633g/d - 124.56g/d = 508.44 g/d The energy needed for vaporization is Q = ?m · ?h_vap = 508.44g/d · 580cal = -294895.2 cal/d = 294.895 kcal/d

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