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Alja [10]
3 years ago
8

The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k=0.029 W/m.K. The measured temperature diff

erence across a 25-mm-thick sheet of the material is LaTeX: T_1-T_2\:=\:12°CT 1 − T 2 = 12 ° C . a) What is the heat flux (in W/m2) through a 3m by 3m sheet of the insulation?
Physics
1 answer:
klasskru [66]3 years ago
3 0

Answer:

flux=13.92W/m^2

Explanation:

The heat transfer by conduction consists in the transport of energy through particles that are together, that is to say by means of solids, Newton developed an equation that allows to know the heat transported in a flat plate knowing the cross sectional area A, Thickness L, conductivity  K  and a temperature difference between the internal and external surface.

Q=KA(t2-t1)/L

To know the heat flux we simply divide both sides of the equation by the area.

Flux=K(t2-t1)/L

where

K=0.029w/mk

L=25mm=0.025m

t2-t1=12 ° C

solving

Flux=(0.029)(12)/(0.025)

flux=13.92W/m^2

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You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re
melamori03 [73]

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

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Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

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Where, C = specific heat capacity of gold

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Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,
REY [17]

Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

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147Sm T ½ = 1.06 1011 years

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238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

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f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0
4 years ago
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