If a person breathes 10 liters per minute of air at 68 degrees Fand 50% relative

Question

If a person breathes 10 liters per minute of air at 68 degrees Fand 50% relative humidity, how much water per day must the internalmembranes supply to saturate the air at 98.6 degrees F? (Assume allthe moisture is exhaled) If each gram of water extracts 580calories 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/m3 and at 37 degreesC is 44.0 g/m cubed.

Hint: 1/2 humidity, 1/2 saturation: inhale 17.3/2= 8.65saturation

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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