1. In an old house, the heating system uses radiators, which are hollow metal de
ID: 1346983 • Letter: 1
Question
1. In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissivity = 0.80). It has a temperature of 55° C. The new owner of the house paints the radiator a lighter color (emissivity = 0.50). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degree Celsius) of the newly painted radiator?
? ° C
2. A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose that a person is unclothed and energy is being lost via radiation from a body surface area of 1.38 m2, which has a temperature of 34° C and an emissivity of 0.700. Also suppose that metabolic processes are producing energy at a rate of 110 J/s. What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature?
? K
Explanation / Answer
heat radiated=emissivity*steffan's constant*temperature^4
where temperature is taken in radian
as heat radiated is kept constant,
then emissivity*temperature^4=constant
==>0.8*(273+55)^4=0.5*T^4
where T is the temperature of newly painted radiator
==>T=368.896 K=(368.896-273) degree celcius=95.896 degree celcius
Q2.
as we know, energy radiated to a colder room by radiation per second
=emissivity*stefan's constant*area*temperature difference^4
let temperature of the coldest room in which person can stand and does not lose
body temeprature is T kelvin.
body temeprature=34 degree celcius=273+34=307 K
then the energy lost by radiation is supploed by metabolic action.
hence 110=0.7*5.67*10^(-8)*1.38*(304^4-T^4)
==>110=5.4772*10^(-8)*(304^4-T^4)
==>304^4-T^4=20.083*10^8
==>T^4=65.324*10^8
==>T=284.3 K
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