Physics Thermal Properties of Matter

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3 weeks ago

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V
Vishal Baghel

Contributor-Level 10

400 * 1 * 12.5 = 500 * 5 * (100 – 36.5)

S = 1 0 6 3 . 5 c a l / g m ° C        

1 g m = 1 6 3 . 5 m o l  

S = 10 cal/mol °C

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V
Vishal Baghel

Contributor-Level 10

d H d t = 2 π x 2 k d T d x

d H d t = 2 π * k * 4 5 1 2 . 5 1 3 = 2 . 2 5 π k W

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Syed Aquib Ur Rahman

Contributor-Level 10

It's the energy that is needed for a change of state to transition from one substance or physical state to another. Remember that the main condition for latent heat is that there is no temperature change when the energy is absorbed or released. 

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Syed Aquib Ur Rahman

Contributor-Level 10

You will be seeing these temperature plateaus appearing on the heating curve at the main melting and boiling points. The reason for that is the energy reaches latent heat. And we know that latent heat is the energy that lets phase transitions occur without increasing the temperature. 

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Vishal Baghel

Contributor-Level 10

r = r? (1 + αΔT) ⇒ r – r? = r? αΔT
⇒ 3 * 10? ³ = 1 * 1.2 * 10? ΔT ⇒ ΔT = 250°C

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alok kumar singh

Contributor-Level 10

 

θ = 3 0 °

A B = R s i n θ = 2 R

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Vishal Baghel

Contributor-Level 10

(K (4A)/l) (100 – θ) = (KA/l) (θ – 50)
⇒ 400 – 4θ = θ – 50 ⇒ 5θ = 450 ⇒ θ = 90°C

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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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New answer posted

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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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