Question

(33%) Problem 3: Warm blooded animals are home otherm that is, they maintain an approximately constant body temperature. (For humans it's about 37°C.) When they are in an environment that is below their optimum temperature, they use energy derived from chemical reactions within their bodies to warm them up One of the ways that animals lose energy to their environment is through radiation. Every object emits electromagnetic radiation that depends on its temperature. For very hot objects like the sun, that radiation is visible light. For cooler objects, like a house or a person, that radiation is in the infrared and is invisible. Nonetheless, it still carries energy. Other ways that energy is lost by a warm animal to a cool environment include conduction (direct touching of a cooler object) and convection (cooler air moving and carrying thermal energy away). See Heat Transfer for a discussion of all three. For this problem, we'l just consider how much energy an animal needs to burn (obtain from internal chemical reactions) in order to stay warm from radiation losses. The rate at which an object loses energy through radiation is given by the Stefan-Boltzmann equation: Rate of energy loss - Acor Where is the absolute (Kelvin) temperature, A is the area of the object, epsion is the emissivity (-1 for a perfect emitter, less for anything else), and is the Stefan-Boltzmann constant 0 = 5.67 * 108 (mK) Consider a patient trying to sleep naked in a cool room (55* For 13°C). Assume that the person being considered is a perfect emter and absorber of radiation (e = 1), has a surface area of about 2.5 m, and a mass of 80 kg. 33% Par(a) A person emits thermal radiation at a rate corresponding to a temperature of 37°C and absorbs radiation at a rate (from the air and walls) corresponding to a temperature of 13° C. Calculate the individual's rate of energy loss due to radition (in Watts Joules/second) w Grade Summary Deductions 0% Potential 100% sin cos) tan 78 9 Submissions cotan asino acoso 5 6 Attempts remaining: 10 atano acotano) sinh 12 3 (4% per attempt) detailed view cosho tanho cotanho 0 • Degrees Radians VO BACKSPACE Submit I give up! Hints: 0 for a 0% deduction. Hints remaining: 0 Feedback: 0% deduction per feedback. 33% Part (b) Assume that the patient produces no energy to keep warm, if they have a specific heat about equal to that of water (1 Cal/kg-°C) how much would their temperature fall in Kelvins in one hour? (1 Callkcal 10 cal). A 33% Part (c) Given that the energy density of fat is 9 Cal/a, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour?

Answer 1

Given:

m = mass of patient = 80 kg

T1 = initial temperature of patient = 37 degree celsius = (37+273) K

T2 = temperature of cool room = 13 degree celsius = (13+273) K

A = surface area of body = $2.5m^{2}$

The body behaves like a perfectly black body.

a) The rate of loss of heat radiation E = $\sigma A(T_{1}^{4}-T_{2}^{4})$

$=5.67*10^{-8}*2.5*((37+273)^{4}-(13+273)^{4})$

$=360.70W$ [answer]

b) The time given is t = 1 hour = 1*60*60 s

given : s = specific heat of water is 1 cal/g/C = 4.2 J/g/C = 4.2 J/g/K = 4200 J/kg/C

T1 = initial temperature = 37 degree celsius = (37+273) K

Let the final temperature be T.

Therfore,

$E*t=ms(T_{1}-T)$

$\Rightarrow T =T_{1}- \frac{E*t}{ms}$

$=(37+273)- \frac{360.70*1*60*60}{80*4200}=306.14 K$ [answer]

c) Given: Energ density of fat = d = 9 cal/g = 9*4.2J/g

Total energy lost in 1 hour is Q = E*t = 360.70*1*60*60 J

Let mass of fat to be utilize in this 1 hr is m.

Therfore,

m*d = Q

$\Rightarrow m = \frac{Q}{d}=\frac{360.70*1*60*60}{9*4.2}=34352.38 g$

Therefore, 34352.38 g of fat would utilize to maintain the temperature. [answer]