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elena-14-01-66 [18.8K]
3 years ago
13

What is the heat flux (W/m^2), due to radiation heat transfer, from a black body if the surface temperature is 1000C? The convec

tion heat transfer coefficient is 55 W/(m^2 C).
Physics
2 answers:
12345 [234]3 years ago
7 0

Answer:

55000 W/m²

Explanation:

Parameters given:

Surface temperature, T = 1000°C

Hear transfer coefficient, h = 55 W/m²C

Convection heat transfer coefficient is given as:

h = Heat flux/Temperature

Hence, Heat Flux, q, is given as:

q = h * T

q = 55 * 1000 = 55000 W/m²C

zepelin [54]3 years ago
6 0

Answer:

55000 W/m²

Explanation:

Using

ψ = Hf/θ.................... Equation 1

Where Ψ = convection heat transfer, Hf = heat flux, θ = Surface Temperature

Make Hf the subject of the equation,

Hf = ψθ.................. Equation 2

Given: Ψ = 55 W/m².°C, θ = 1000°C

Substitute into equation 2

Hf = 55(1000)

Hf = 55000 W/m²

Hence the heat flux due to radiation = 55000 W/m²

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6 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

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C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

5 0
3 years ago
What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a
Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

A = W + \dfrac{W}{2}

A =\dfrac{3W}{2}

A =\dfrac{3mg}{2}

When the car is at the bottom,  the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

N = m g + \dfrac{mv^2}{r}

\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}

\dfrac{mg}{2} = \dfrac{mv^2}{r}

v = \sqrt{\dfrac{rg}{2}}

v = \sqrt{\dfrac{40\times 9.8}{2}}

v = 14 m/s

8 0
3 years ago
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