Question

A cylindrical log 15cm in diameter and 68cm long is glowing red hot in a fireplace
If it 's emitting radiation at the rate of 38kW, what is its temperature? the log's emissivity e=1
Please , help

Answer 1

\[\large T=\sqrt[4]{\frac{P}{e\sigma A}}\]
where P = 3.8*10^4 W
e =1
A = \((\dfrac{0.15}{2})^2 * 0.68 = 0.003825 m^2 = 0.3825 *10^{-2}\)
am I right?

Answer 2

@LastDayWork

Answer 3

@agent0smith

Answer 4

oh, I calculate A without time to pi :) , now redo :) A = 1.20 x 10^(-2) m^2

Answer 5

Are you including the areas of both the ends and of the round portion?

Answer 6

My area came out to be 0.782256571 m^2

Answer 7

my calculator said it is 0.355785

Answer 8

What formula are you using to calculate area ?? (in terms of r and h)

Answer 9

area bottom and top are = 2 (r^2 *pi) where r = 0.15/2 = 0.075
area of the side is = 2rpi *h

Answer 10

Okay..then A=0.355785368 m^2 is correct.
I only looked at the diagram and got the impression r = 0.15 m XD

Answer 11

Then every thing else is correct; you only need to plug-in the values :)

Answer 12

thank you @LastDayWork
I have another question from the concept
An 18m by 4.5 m house built on a concrete slab 27 cm thick
What is the heat-loss rate throught the floor if the interior is at 29 C while the ground is at 19 C, thermal conductivity of concrete is 1W/m.K
I know \(H= -kA (T_f - T_i) in this case, how to define which T is T_f?

Answer 13

and since it is the heat lost (interior temperature > the floor temperature) so, H will have - sign, right?

Answer 14

Need thickness x there, too


heat loss = - k A (T1-T2)/x

Answer 15

oh yes, sorry, I missed it. I calculate and get -3000W . they ask me to write answer using 2 sig fig. is it 30.00 x 10^2 or -30.00 x10^2