Thermal Properties of Matter Overview

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V
Vishal Baghel

Contributor-Level 10

Δ V L = V 0 γ Δ T

Δ V s = V 0 * 3 * γ 3 * Δ T

Δ V L Δ V S = 0

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Vishal Baghel

Contributor-Level 10

Thermal stress is developed on heating when expansion of rod is hindered.

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Vishal Baghel

Contributor-Level 10

According to question, we can write

d = d 0 ( 1 + α Δ T ) 6 . 2 4 1 = 6 . 2 3 0 ( 1 + 1 . 4 * 1 0 5 * Δ T )

T = 1 2 6 . 1 8 + 2 7 = 1 5 3 . 1 8 ° C  

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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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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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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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a month ago

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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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