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3 years ago

Round the following answer correctly. 5.40m x 3.21m x 1.871m = 32.431914 m^3

1 answer:

Klio2033 [76]3 years ago

5 0

The answer is 32.43 because when you round the number next to the hunthers place (witch in this problem would be the 1) has to be 5 or hight to raise the number to the left of it and if it is lower you just take the two number and that is your answer.

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3 years ago

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3 years ago

What are the principal energy level? And what do they represent?

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3 years ago

Describe in your own words, in terms of particle movement and energy, what occurs when a liquid is heated to its boiling point.

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3 years ago

The wall of an industrial furnace is constructed from 0.15-m-thick, fireclay brick having a thermal conductivity of LZ W/m.K. Me

aliya0001 [1]

Answer:

The rate of heat loss through the wall is 1700 watts.

Explanation:

The complete statement of the problem is:

<em>The wall of an industrial furnace is constructed from 0.15-m-thick, fireclay brick having a thermal conductivity of 1.7 W/m·K. Measurements made during steady state operation reveal temperatures of 1400 and 1150 K at the inner and outer sur- faces, respectively. What is the rate of heat loss through a wall that is 0.5 m by 1.2 m on a side?</em>

Given that wall of the industrial furnace is under steady conditions of heat transfer and whose configuration is a flat element, we use the equation of conductive heat transfer rate ( $\dot Q$ ), measured in watts:

$\dot Q = \frac{k\cdot w\cdot h}{l}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Kelvin.

$w$ - Width of the wall, measured in meters.

$h$ - Height of the wall, measured in meters.

$l$ - Thickness of the wall, measured in meters.

$T_{i}$ - Inner surface temperature, measured in Kelvin.

$T_{o}$ - Outer surface temperature, measured in Kelvin.

If we know that $k = 1.7 \,\frac{W}{m\cdot K}$ , $w = 1.2\,m$ , $h = 0.5\,m$ , $l = 0.15\,m$ , $T_{i} = 1400\,K$ and $T_{o} = 1150\,K$ , the steady state heat transfer is:

$\dot Q = \left[\frac{\left(1.7\,\frac{W}{m\cdot K} \right)\cdot (1.2\,m)\cdot (0.5\,m)}{0.15\,m} \right]\cdot (1400\,K-1150\,K)$

$\dot Q = 1700\,W$

The rate of heat loss through the wall is 1700 watts.

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3 years ago