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Sphinxa [80]
3 years ago
10

In an electric furnace used for refining steel, the temperature is monitored by measuring the radiant power emitted through a sm

all hole in the wall of the furnace, of area 0.5 cm2. This hole acts like a perfect blackbody radiator having the same temperature as the interior of the furnace. If the temperature of the furnace (and therefore of the hole) is to be maintained at 1650°C, how much power will the hole radiate?
Physics
1 answer:
stiks02 [169]3 years ago
3 0

Answer: 3.7×10¹²watts

Explanation:

Radiation is one of the mode of heat transfer and modes differs from each other based on their medium of heat transfer. Radiation is a process of transferring heat energy from one point to another without heating the intervening medium (no material medium is required).

According to Stefan's law of radiation, the rate of emission of radiant energy is directly proportional to the fourth power of its absolute temperature.

Mathematically, R = eAT⁴

e is constant of proportionality called emissivity. Emissivity varies depending on the type of body being considered.

For the question, we are considering black body and emissivity of black body is 1 being a perfect body.

A is the area of the body

T is the absolute temperature

e = 1

A = 0.5cm²

T = 1650°C

Rate of radiation = 1×0.5×1650⁴

= 3.7×10¹²watts.

The hole will therefore radiate 3.7×10¹²watts

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On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3, length 88.8 cm and diameter 2.30 cm fro
vichka [17]

Answer:

w = 28.25 N

Explanation:

To do this, we need to use two expressions.

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w = m*g  (1)

However we do not have the mass of the rod. We need to calculate that. To calculate the mass, we'll use the expression of density which is:

d = m/V  

From here, we solve for mass:

m = d * V   (2)

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V = π * r² * h   (3)

So, in resume, we need to solve for the volume of the rod, then, the mass ans finally the weight. Let's calculate the volume of the rod, converting the units of centimeter to meters, just dividing by 100:

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Replacing in (3):

V = π * (0.0115)² * 0.888

V = 3.69x10⁻⁴ m

Now, let's use (2) to calculate the mass:

m = 7800 * 3.69x10⁻⁴

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Finally for the weight, we'll use expression (1):

w = 2.88 * 9.81

<h2>w = 28.25 N</h2><h2>And this is the weight of the rod.</h2>
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