Suppose a person metabolizes 2250 kcal/day.
a) With a core body temperature of 36.9°C and an ambient temperature of 21.5°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 1.59 m2, a skin temperature of 27.1°C, and an effective emissivity of e = 0.650, at what net rate does this person radiate heat to the 21.5°C surroundings?
(in W)
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 36.9°C.
(in W/K)
To solve these questions, we can use various formulas and concepts related to thermodynamics and heat transfer. Let's go through each question step by step.
a) The maximum efficiency of a heat engine is given by the Carnot efficiency. It is defined as the ratio of the work output to the heat input. The formula for Carnot efficiency is:
Efficiency (η) = 1 - (Tc/Th)
where Tc is the temperature of the cold reservoir (in Kelvin) and Th is the temperature of the hot reservoir (in Kelvin).
In this case, we are given the core body temperature (Tc) as 36.9°C and the ambient temperature (Th) as 21.5°C. We need to convert these temperatures to Kelvin before applying the formula.
Tc = 36.9 + 273.15 = 310.05 K
Th = 21.5 + 273.15 = 294.65 K
Now we can plug the values into the formula to find the Carnot efficiency.
Efficiency (η) = 1 - (310.05/294.65) = 0.0476 or 4.76% (rounded to two decimal places)
Therefore, the maximum Carnot efficiency with which the person can perform work is approximately 4.76%.
b) The rate at which the person has to shed waste heat to the surroundings can be calculated using the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted from one form to another.
The formula for calculating heat transfer is:
Heat transfer (Q) = Efficiency (η) * Metabolic rate (M)
Here, the metabolic rate is given as 2250 kcal/day. To convert it to watts, we need to multiply it by a conversion factor:
1 kcal/day = 1.163 W
So, the metabolic rate in watts is:
M = 2250 * 1.163 = 2615.25 W
Now we can calculate the heat transfer:
Q = 0.0476 * 2615.25 = 124.539 W
Therefore, the person would have to shed waste heat to the surroundings at a rate of approximately 124.54 watts.
c) To calculate the rate at which the person radiates heat to the surroundings, we can use the Stefan-Boltzmann law. It states that the power radiated by an object is proportional to the fourth power of its temperature and its effective emissivity.
The formula for radiated heat transfer is:
Heat transfer (Q) = σ * A * e * [(T1^4) - (T2^4)]
where σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2K^4)), A is the surface area, e is the effective emissivity, T1 is the body temperature, and T2 is the ambient temperature.
Given the surface area (A) as 1.59 m^2, body temperature (T1) as 27.1°C, and ambient temperature (T2) as 21.5°C, we need to convert these temperatures to Kelvin.
T1 = 27.1 + 273.15 = 300.25 K
T2 = 21.5 + 273.15 = 294.65 K
Now we can plug the values into the formula:
Q = (5.67 x 10^-8) * 1.59 * 0.65 * [(300.25^4) - (294.65^4)]
Calculating this expression will give you the net rate at which the person radiates heat to the 21.5°C surroundings.
d) The rate at which the person loses water can be calculated using the latent heat of vaporization of water. This is the amount of heat required to convert a substance from a liquid to a gas phase without changing its temperature.
The formula for calculating the rate of water loss is:
Rate of water loss (W) = Waste heat transfer (Q) / Latent heat of vaporization of water (L)
Given that the latent heat of vaporization of water is 575 cal/g, we need to convert it to watts by using the conversion factor:
1 cal = 4.184 J
So, the latent heat of vaporization of water in watts is:
L = 575 * 4.184 = 2402.6 J/g
Now we can calculate the rate of water loss:
W = Q / L
You can use the value of Q calculated in part b) and divide it by L to obtain the rate of water loss in grams per hour.
e) To estimate the rate at which the person gains entropy, we can use the formula:
Rate of entropy gain (S) = Waste heat transfer (Q) / Temperature (T)
Given that the waste heat transfer (Q) is calculated in part b) and the temperature (T) is the core body temperature at 36.9°C, convert it to Kelvin:
T = 36.9 + 273.15 = 310.05 K
Now we can calculate the rate of entropy gain:
S = Q / T
This will give you the rate at which the person gains entropy in wattsper Kelvin (W/K).