For this problem, we\'ll just consider how much energy an animal needs to burn (

Question

For this problem, we'll just consider how much energy an animal needs to burn (obtain from internal chemical reactions) in order to stay warm just from radiation losses. The rate at which an object loses energy through radiation is given by the Stefan-Boltzmann equation:

Rate of energy loss = AT4

where T is the absolute (Kelvin) temperature, A is the area of the object, is the emissivity (unitless and =1 for a perfect emitter, less for anything else), and is the Stefan-Boltzmann constant:

= 5.67 x 10-8 J/(s m2 K4)

Consider a patient trying to sleep naked in a cool room (55 oF = 13 oC). Assume that the person being considered is a perfect emitter and absorber of radiation ( = 1), has a surface area of about 2.5 m2, and a mass of 80 kg.

a. A person emits thermal radiation at a rate corresponding to a temperature of 37 oC and absorbs radiation at a rate (from the air and walls) corresponding to a temperature of 13 oC. Calculate the individual's net rate of energy loss due to radiation (in Watts = Joules/second). ___________

b. Assume the patient produces no energy to keep warm. If they have a specific heat about equal to that of water (1 Cal/kg-oC) how much would their temperature fall in one hour? (1 Cal = 1kcal = 103 cal)
T = ______ oC

c. Given that the energy density of fat is about 9 Cal/g, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour?
amount of fat needed = ______g

Explanation / Answer

(a)

The net rate of energy is

Q(37°C) - Q (13°C) = A(37°C+273)^4-A(13+273)^4= A(37-13)^4=0.047 J/s

(b)

- Q (13°C) *t=m*cp*T

t=1 hour=3600 s

Cp=4.18 J/kg-°C

T =3600*(-2.5* 5.67 x 10-8 * (13^4))/(80*4.18)= 0.043 °C

(c)

- Q (13°C) *t=m*cp*T

T =37-13 =24°C

m= ( Q (37°C)- Q (13°C)) *t/ (cp*(24))

cp=37.62 J/kg-K and t=3600 s

m=0.17 kg=170 g

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