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 Post subject: Heat RadiationPosted: Wed, 14 Dec 2005 01:09:26 UTC

Joined: Sun, 4 Dec 2005 15:33:45 UTC
Posts: 7
Any object at Kelvin temperature T will radiate electromagnetic waves, typically infrared radiation for bodies at room temperature with some visible light for bodies heated above 1000 K. The spectrum of this blackbody radiation depends only on the temperature and not on the type of material. The formula governing the total power radiated is P = e(sigma)AT^4 where sigma is Stefan-Boltzmann constant

1. this formula is applied to:
A. any object of total surface area A; Kelvin temperature T; and emissivity e
B. any object of cross-sectional area A; Kelvin temperature T; and emissivity e
C. any object of total surface area A; Kelvin temperature T; and emissivity sigma
D. any object of cross-sectional area A; Kelvin temperature T; and emissivity sigma

*I think it's A.

2. If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from the color of the light), the power input, the Stefan-Boltzmann constant, and what property of the filament?

B. emissivity
C. length

*I think it's B.

3. If you calculate the thermal power radiated by typical objects at room temperature, you will find surprisingly large values--several kilowatts typically. For example, a square box that is 1 m on each side and painted black (therefore justifying an emissivity e near unity) emits 2.5 kW at a temperature of 20 Celsius. In reality the net thermal power emitted by such a box must be much smaller than this, or else the box would cool off quite quickly. Which of the following alternatives seems to best explain this conundrum?

A. The box is black only in the visible; in the infrared (where it radiates) it is quite shiny and radiates little power.
B. The surrounding room is near the temperature of the box and radiates about 2.5 kW of thermal energy into the box.
C. Both of the first two factors contribute significantly.
D. Neither of the first two factors is the explanation.

*I think it's B.

4. As a rough approximation, the human body may be considered to be a cylinder of length L = 2.0 m and circumference C = 0.8 m. (To simplify things, ignore circular top and bottom and just consider the cylindrical sides.) If the emissivity of the skin is taken to be e = 0.6 and its surface temperature is taken to be T = 30 Celsius, how much thermal power P does the human body radiate?
Express the power radiated numerically; give your answer to the nearest 10 W.

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 Post subject: Posted: Wed, 14 Dec 2005 02:23:42 UTC
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Joined: Mon, 23 Jun 2003 17:34:53 UTC
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Location: San Antonio,Texas USA
4. The area of a cylinder is , and .

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 Post subject: Posted: Thu, 15 Dec 2005 13:20:53 UTC

Joined: Sun, 4 Dec 2005 15:33:45 UTC
Posts: 7
A = LC = 1.6
T = 303 K
e = 0.6

P = 1.6*0.6*303^4 = 8.091737*10^9 W

ans to the nearest 10 W is that 8.1*10^9?

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 Post subject: Posted: Thu, 15 Dec 2005 13:43:45 UTC
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Joined: Mon, 23 Jun 2003 17:34:53 UTC
Posts: 2094
Location: San Antonio,Texas USA
Looks like you forgot to multiply by .

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 Post subject: Posted: Sat, 24 Dec 2005 00:26:41 UTC

Joined: Sun, 4 Dec 2005 15:33:45 UTC
Posts: 7
what is sigma?

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 Post subject: Posted: Sat, 24 Dec 2005 06:27:53 UTC
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Joined: Wed, 2 Mar 2005 07:44:02 UTC
Posts: 593
Location: Parsippany, NJ
You defined it before as the Stefan-Boltzmann constant. Read your first post.

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