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SS1: PHYSICS – 1ST TERM

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  1. Introduction to Physics | Week 1
    4 Topics
    |
    1 Quiz
  2. Measurement I | Week 2
    3 Topics
    |
    1 Quiz
  3. Measurement II | Week 3
    6 Topics
    |
    1 Quiz
  4. Motion | Week 4
    5 Topics
    |
    1 Quiz
  5. Velocity-Time Graph | Week 5
    4 Topics
    |
    1 Quiz
  6. Causes of Motion | Week 6
    5 Topics
    |
    1 Quiz
  7. Work, Energy & Power | Week 7
    3 Topics
  8. Energy Transformation / Power | Week 8
    3 Topics
    |
    1 Quiz
  9. Heat Energy | Week 9
    5 Topics
    |
    1 Quiz
  10. Linear Expansion | Week 10
    7 Topics
    |
    1 Quiz
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Topic Content:

  • Definition of Area Expansivity
  • Calculations on Area Expansivity

When substances are heated, there is an increase in length, breadth and height, hence area and volume also increases.

An increase in area of a solid when heated is called area or superficial expansivity.

Therefore;

Area Expansivity = β

\(\scriptsize (β) = \normalsize \frac {Increase \: in \: Area}{Original \: Area \: \times \: Temperature \: Change}\)

 \(\scriptsize β = \normalsize \frac {A_2 \: – \: A_1}{A_1 \: \times \: (\theta_2 \: – \: \theta_1)}\) 

A2 = New area

A1 = Initial/Original area

θ2 = New temperature

θ1 = Original temperature

From the equation

A2 – A1 =  β A1( θ2 –  θ1)

A2 = A1 +  β A1( θ2 – θ1)

or

A2 = A1 (1 +  β(θ2 – θ1))

Area expansivity = 2 * Linear expansivity

β = 2α ( Proof of The Relationship between Linear Expansivity and Area Expansivity)

The S.I unit is per kelvin \( \scriptsize K^{-1} \) or per degree Celsius \( \scriptsize ^{०} C^{-1} \)

Example 10.3.1:

Calculate the area expansivity of a metal plate whose area at 36°C is 0.25 m2 and 1.15 m2 at 56.2 °C.

Solution

Values given:

Area expansivity β = ?, A1 = 0.25m2 , A2 = 1.15m2 , θ1 = 36°C, θ2 = 56.2°C

Formula for area expansivity β

\(\scriptsize β = \normalsize \frac {A_2 \: – \: A_1}{A_1 \: \times \: (\theta_2 \: – \: \theta_1)}\) 

 ⇒ \(\scriptsize β = \normalsize \frac {1.15\: – \: 0.25}{0.25 \: \times \: (56.2 \: – \: 36)}\) 

 ⇒ \(\scriptsize β = \normalsize \frac {0.9}{0.25 \: \times \: 20.2}\) 

⇒ \(\scriptsize β = \normalsize \frac {0.9}{5.05}\) 

⇒ \(\scriptsize β = 0.178 K^{-1}\) 

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