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DENIUS [597]
3 years ago
11

How much can having a lighted candle increase the temperature inside the shelter?

Physics
1 answer:
oksian1 [2.3K]3 years ago
8 0
A lighted candle produces heat however not as much heat as a heater or the sun would.
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If you were told an atom was an ion, you would know the atom must have a
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The atom must have gained 1 or more electrons or must have lost 1 or more electrons.
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Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

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3 years ago
A change in which property of light will have no effect on whether or not the photoelectric effect occurs??? Frequently energy i
MaRussiya [10]
The intensity of the light has no connection with the photoelectric effect.

That's what was so baffling about it before the particle nature of light
was suspected ... a match with a blue flame might stimulate the
photoelectric effect, but a high-power red searchlight couldn't do it.
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I need help with number 2 and 3
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1. Gas
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What eccentricity value results in a circular orbit?
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Answer:

Zero

Explanation:

Given the equation of an ellipse:

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Therefore the eccentricity of a circle is

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