Question

(33%) Problem 3: Warm blooded animals are homeothermic 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'll 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 Acar Where Tis the absolute (Kelvin) temperature, A is the area of the object, epsilon is the emissibilty (= 1 for a perfect emitter, less for anything else), and is the Stefan Boltzmann constant 0 = 5.67 * 10% (m2K) Consider a patient trying to sleep naked in a cool room (55° F or 13° C). Assume that the person being considered is a perfect emiter and absorber of radiation (e = 1), has a surface area of about 2.5 m², and a mass of 80 kq. 33% Part(a) A person emits thermal radiation at a rate corresponding to a temperature of 3? Cand 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). P-360.7 ✓ Correct! * 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 Cal = 1kcal = 10 cal). AT-292.14 AT-292.1 X Attempts Remain * 33% Part (c) Given that the energy density of fat is ng Cal/a, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour? m= 34352.41 Grade Summary g Deductions 3% Potential 92% sino coso tano 78 9 HOME Submissions colano asino acos E14 5 6 Attempts remaining: 8 atan acotano) sinh ) 1 12 3 (4% per attempt) detailed view cosho tanho cotanho 0 END 4% • Degrees Radians VO BACKSPACE CLEAR 4% + 2

Answer 1

Find the rate of energy loss using the area and temperature of body and room as shown below. Use the specific heat after finding the energy lost in 1 hour to find the decrease in temperature

- а 1 Rate = P = A6(74-74) {emissivity = (2.5m2) (5-67% 10% 5/8 m2**)) (273+3774 - (273+2374) 1. 360:71 W Il = = specific heat S 1 kcall 19.00 4184 J 5/1gic Heat lost in in 1 hour is, Os P t = (360.7 W) (36005) - 1.298x101 Now, B = MS AT = 1.298x1065 = (8019) 41841/192) AT 5 AT = 3.88K lg gives E - q cal = (1000 call xq (4184 J) - So, for B= 1.298 X 1065 m il 8 olui 1:298X1065 9x4184J/ 11 34.5 g oo