Physics Thermal Properties of Matter

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

V = 2 α 2 + α 1

= 10 * 10 - 6 + 5 * 10 - 5

= 60 * 10 - 6 / ? C

60

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2 months ago

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A
alok kumar singh

Contributor-Level 10

M ice   L f + m ice   ( 40 - 0 ) C w = m steam   L v + m steam   ( 100 - 40 ) C w

200 [ 80 + 40 ( 1 ) ] = M [ 540 + 60 ( 1 ) ]

200 ( 120 ) = M ( 600 )

M = 40 g m .

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

l B ( 1 + α B Δ T ) l i ( 1 + α i Δ T ) = l B l i

α B l B = l i α i

l i = 6 0 c m

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Vishal Baghel

Contributor-Level 10

At steady state condition -> Heat conducted through slab AB = Heat conducted though slab BC

k 1 A A B ( 1 0 0 8 0 ) = k A B C R ( 8 0 0 )

-> k 1 1 6 * 2 0 = k 8 * 8 0

k1 = 8 k

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alok kumar singh

Contributor-Level 10

keq= L 1 + L 2 L 1 k 1 + L 2 k 2 = 4 + 2 . 5 4 k + 2 . 5 2 k

keq= 6 . 5 8 + 2 . 5 2 k = 6 . 5 1 0 . 5 * 2 k = 1 3 * 2 2 1 k

= ( 1 + 5 5 1 ) k

= ( 1 + 5 α ) =21

New answer posted

2 months ago

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alok kumar singh

Contributor-Level 10

Area=2* [lb+bh+hl]

A=2* [0.6*0.5+0.5*0.2+0.2*0.6]

= 1.04 M2

Rthermal=tKA=1*1020.05*1.04

m=61*105kg/s

 

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