Suppose a person metabolizes 2150 kcal/day.
a) With a core body temperature of 37.1°C and an ambient temperature of 21.3°C, what is the maximum (Carnot) efficiency with which the person can perform work?
b) If the person could work with that efficiency, at what rate, in watts, would they have to shed waste heat to the surroundings?
c) With a skin area of 1.51 m2, a skin temperature of 26.7°C, and an effective emissivity of e = 0.610, at what net rate does this person radiate heat to the 21.3°C surroundings?
d) The rest of the waste heat must be removed by evaporating water, either as perspiration or from the lungs. At body temperature, the latent heat of vaporization of water is 575 cal/g. At what rate, in grams per hour, does this person lose water?
e) Estimate the rate at which the person gains entropy. Assume that all the required evaporation of water takes place in the lungs, at the core body temperature of 37.1°C.
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a) To determine the maximum efficiency at which the person can perform work, we can use the Carnot efficiency formula, which is given by:
Efficiency = 1 - (Tc/Th)
Where Tc is the temperature of the cold reservoir (ambient temperature) and Th is the temperature of the hot reservoir (core body temperature). In this case, Tc = 21.3°C and Th = 37.1°C.
Plugging in the values into the formula:
Efficiency = 1 - (21.3/37.1) = 0.426 or 42.6%
Therefore, the maximum efficiency with which the person can perform work is 42.6%.
b) To calculate the rate at which the person would have to shed waste heat to the surroundings, we can use the formula for power, given by:
Power = Efficiency * Metabolic Rate
Metabolic Rate represents the amount of energy the person metabolizes per unit time (given as 2150 kcal/day). To convert kcal/day into watts, we use the conversion factor: 1 kcal/day = 0.0116 W.
Power = Efficiency * Metabolic Rate = 0.426 * (2150 kcal/day * 0.0116 W/kcal/day) = 102.4 W
Therefore, the person would have to shed waste heat to the surroundings at a rate of 102.4 watts.
c) To calculate the net rate at which the person radiates heat to the surroundings, we use the Stefan-Boltzmann law, given by:
Power = σ * A * e * (T^4 - Tc^4)
Where σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m^2K^4), A is the skin area (1.51 m^2), e is the effective emissivity (0.61), T is the skin temperature (26.7°C + 273.15 K), and Tc is the ambient temperature (21.3°C + 273.15 K).
Plugging in the values into the formula:
Power = σ * A * e * (T^4 - Tc^4)
Power = (5.67 x 10^-8) * 1.51 * 0.61 * ((26.7 + 273.15)^4 - (21.3 + 273.15)^4)
Power ≈ 101.2 W
Therefore, the person radiates heat to the 21.3°C surroundings at a net rate of approximately 101.2 watts.
d) To determine the rate at which the person loses water through evaporation, we need to calculate the waste heat that is removed by evaporating water. We know that the latent heat of vaporization of water at body temperature is 575 cal/g.
First, we convert the metabolic rate from watts to calories per hour:
Metabolic Rate = 2150 kcal/day
Metabolic Rate = 2150 kcal/24 hours
Metabolic Rate = 89.6 kcal/hour
Next, we divide the metabolic rate by the latent heat of vaporization of water to find the rate of water loss:
Water Loss Rate = Metabolic Rate / Latent Heat of Vaporization
Water Loss Rate = 89.6 kcal/hour / 575 cal/g
To convert kcal to grams, we use the conversion factor: 1 kcal = 1000 cal.
Water Loss Rate = 89.6 kcal/hour / 575 cal/g * 1000 g/kg
Water Loss Rate ≈ 0.155 kg/hour
Therefore, the person loses water at a rate of approximately 0.155 kilograms per hour.
e) To estimate the rate at which the person gains entropy, we need to consider the heat transfer due to evaporation. The rate of entropy gain is given by:
Entropy Gain Rate = Waste Heat / Temperature
Since all required evaporation takes place in the lungs at the core body temperature of 37.1°C, we can use this temperature.
Waste Heat = Metabolic Rate - Power Radiated
Waste Heat = 89.6 kcal/hour - 101.2 W
Converting kcal to watts using the conversion factor: 1 kcal/hour = 0.0116 W:
Waste Heat = 89.6 kcal/hour * 0.0116 W/kcal/hour - 101.2 W
Waste Heat ≈ 1.04 W
Entropy Gain Rate = Waste Heat / Temperature = 1.04 W / 37.1°C ≈ 0.028 J/K
Therefore, the estimated rate at which the person gains entropy is approximately 0.028 J/K.