Physics Thermal Properties of Matter

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2 months ago

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A
alok kumar singh

Contributor-Level 10

  Δ L = L 0 α Δ T

= 12 * 11 * 10-6 * 30

= 3960 * 10-6 m

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A
alok kumar singh

Contributor-Level 10

γ = 3 α

Δ V = γ V Δ T = 3 α . a 3 . Δ T

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V
Vishal Baghel

Contributor-Level 10

R_eq = R? + R?

L_total / (K_eq * A) = L? / (K? A) + L? / (K? A)
Assuming L? = L? = l, L_total = 2l

2l / (K_eq * A) = l/ (K? A) + l/ (K? A)

2/K_eq = 1/K? + 1/K? = (K? + K? )/ (K? )

K_eq = 2K? K? / (K? + K? )

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A
alok kumar singh

Contributor-Level 10

In steady state rate of flow of heat in all three rods are same.
dQ/dt = (k? A (100 – 70)/? = (k? A (70 – 20)/? = (k? A (20 – 0)/?
30k? = 50k? = 20k?
∴ k? : k? = 2:3 & k? : k? = 2:5

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New answer posted

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Vishal Baghel

Contributor-Level 10

At T°C L = L? + L?
At T +? T Leq = L'? + L'?
where L'? = L? (1 + α? T)
L'? = L? (1 + α? T)
Leq = (L? + L? ) (1 + αavg? T)


⇒ (L? + L? ) (1 + αavg? T) = L? + L? + L? α? T + L? α? T
⇒ (L? + L? )αavg = L? α? + L? α?
⇒ αavg = (L? α? + L? α? )/ (L? + L? )

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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

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