Physics Worksheet 4 - Grade 9 - Temperature and Thermal Expansion Notes

Temperature

  • Temperature is the measure of hotness or coldness of a body.

  • Temperature is a measure of the average kinetic energy of the particles of a body.

  • Temperature is an intensive property and a fundamental quantity.

  • A body with particles having large kinetic energy will have a high temperature.

  • A thermometer is used to measure temperature.

  • The SI unit of temperature is Kelvin (K).

  • Joule is NOT a unit of temperature; it's a unit of energy.

  • Temperature scales include Centigrade (Celsius), Fahrenheit, and Kelvin.

Temperature Scales

  • Celsius Scale:

    • Lower fixed point (ice point): 0 °C
    • Upper fixed point (steam point): 100 °C

  • Fahrenheit Scale:

    • Lower fixed point (ice point): 32 °F
    • Upper fixed point (steam point): 212 °F

  • Kelvin Scale:

    • Lower fixed point (ice point): 273 K
    • Upper fixed point (steam point): 373 K

Temperature Conversions

  • Celsius to Fahrenheit: F=(9/5)C+32F = (9/5)C + 32

  • Fahrenheit to Celsius: C=(5/9)(F−32)C = (5/9)(F - 32)

  • Celsius to Kelvin: K=C+273.15K = C + 273.15

  • Kelvin to Celsius: C=K−273.15C = K - 273.15

  • Average human body temperature is 37 °C, which is 98.6 °F.

  • Boiling point of water:

    • 100 °C
    • 212 °F
    • 373 K

  • On a hot day, the Fahrenheit scale shows the greatest number.

  • A change in temperature of 55 °C is equal to a change of 55 K.

  • Celsius and Fahrenheit scales read the same at -40 degrees: −40°C=−40°F-40 °C = -40 °F

  • 20 °C is equal to 293 K.

  • 20°C is equivalent to 68°F.

  • 60 °C is equivalent to 140 °F.

  • 310 K is equivalent to 37 °C.

  • 86 °F is equivalent to 30 °C.

  • 273 K is equivalent to 0 °C.

  • 77 °F is approximately 298 K.

  • The set of equivalent temperatures: 50 °F, 10 °C, 283 K.

  • 273 K is equal to 32 °F.

  • 25 °C is equal to 77 °F and 298 K.

Heat Flow

  • Temperature determines the direction of heat flow.

  • -40 °F is equal to 233 K.

Thermal Expansion

  • When the temperature of a thin metal rod increases, its length also increases.

  • Change in length is directly proportional to the change in temperature.

    • ΔL=αL<em>0ΔT\Delta L = \alpha L<em>0 \Delta T where α\alpha is the coefficient of linear expansion, L</em>0L</em>0 is the original length, and ΔT\Delta T is the change in temperature

  • States of matter arranged from lower to higher degree of expansion: solid, liquid, gas.

  • When a metal ball is heated, it expands and cannot pass through a ring.

Workout Problems

  1. 0 °F is approximately -17.8 °C.

  2. If the ice and steam points are 250X and 1250X respectively, and a temperature is 62 °C, the corresponding reading on this thermometer "X" needs to be calculated using the ratio X−2501250−250=C−0100−0\frac{X - 250}{1250 - 250} = \frac{C - 0}{100 - 0}, where C = 62

  3. A Celsius degree represents a larger temperature change than a Fahrenheit degree.

  4. Relationship between Celsius and Fahrenheit temperature changes: ΔT<em>F=95ΔT</em>C\Delta T<em>F = \frac{9}{5} \Delta T</em>C

  5. Heating water from 25°C to 80°C:

    • Change in Kelvin scale: 55 K
    • Change in Fahrenheit scale: 99 °F

  6. Steel railroad track length at 40°C:

    • Given: Initial length L<em>0=30L<em>0 = 30 m at 0°C, ΔT=40\Delta T = 40°C, and the linear expansion coefficient of steel α</em>steel≈12×10−6 °C−1\alpha</em>{steel} \approx 12 \times 10^{-6} \text{ °C}^{-1}.
    • Using the linear expansion formula: ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T
    • ΔL=(12×10−6 °C−1)(30 m)(40 °C)=0.0144 m=1.44 cm\Delta L = (12 \times 10^{-6} \text{ °C}^{-1}) (30 \text{ m}) (40 \text{ °C}) = 0.0144 \text{ m} = 1.44 \text{ cm}
    • The length at 40°C is L=L0+ΔL=30 m+0.0144 m=30.0144 mL = L_0 + \Delta L = 30 \text{ m} + 0.0144 \text{ m} = 30.0144 \text{ m}

  7. Determining the temperature at which a steel bar and a brass bar will be the same length:

    • Given: Steel bar initial length L<em>s0=3 mL<em>{s0} = 3 \text{ m} at 20°C, brass bar initial length L</em>b0=2.98 mL</em>{b0} = 2.98 \text{ m} at 20°C.
    • Linear expansion coefficients: α<em>steel≈12×10−6 °C−1\alpha<em>{steel} \approx 12 \times 10^{-6} \text{ °C}^{-1}, α</em>brass≈19×10−6 °C−1\alpha</em>{brass} \approx 19 \times 10^{-6} \text{ °C}^{-1}.
    • Let TT be the temperature at which the lengths are equal. Then:
    • L<em>s=L</em>s0(1+α<em>steel(T−20))L<em>s = L</em>{s0} (1 + \alpha<em>{steel} (T - 20)) and L</em>b=L<em>b0(1+α</em>brass(T−20))L</em>b = L<em>{b0} (1 + \alpha</em>{brass} (T - 20))
    • Setting L<em>s=L</em>bL<em>s = L</em>b:
    • 3(1+12×10−6(T−20))=2.98(1+19×10−6(T−20))3 (1 + 12 \times 10^{-6} (T - 20)) = 2.98 (1 + 19 \times 10^{-6} (T - 20))
    • Solving for TT:
    • 3+36×10−6(T−20)=2.98+56.62×10−6(T−20)3 + 36 \times 10^{-6} (T - 20) = 2.98 + 56.62 \times 10^{-6} (T - 20)
    • 0.02=20.62×10−6(T−20)0.02 = 20.62 \times 10^{-6} (T - 20)
    • T−20=0.0220.62×10−6≈969.93T - 20 = \frac{0.02}{20.62 \times 10^{-6}} \approx 969.93
    • T≈989.93 °CT \approx 989.93 \text{ °C}

  8. Change in length of a 520m steel bridge between -20°C and 35°C:

    • Given: Initial length L<em>0=520 mL<em>0 = 520 \text{ m}, ΔT=35−(−20)=55 °C\Delta T = 35 - (-20) = 55 \text{ °C}, and α</em>steel≈12×10−6 °C−1\alpha</em>{steel} \approx 12 \times 10^{-6} \text{ °C}^{-1}.
    • Using the linear expansion formula: ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T
    • ΔL=(12×10−6 °C−1)(520 m)(55 °C)=0.3432 m=34.32 cm\Delta L = (12 \times 10^{-6} \text{ °C}^{-1}) (520 \text{ m}) (55 \text{ °C}) = 0.3432 \text{ m} = 34.32 \text{ cm}

  9. A 50m metal pipe lengthens by 2.85 cm when heated from 10 °C to 40 °C:

    • Given: L0=50 mL_0 = 50 \text{ m}, ΔL=2.85 cm=0.0285 m\Delta L = 2.85 \text{ cm} = 0.0285 \text{ m}, ΔT=40−10=30 °C\Delta T = 40 - 10 = 30 \text{ °C}. a. Determine the linear expansion coefficient:
      • Using the linear expansion formula: ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T
      • α=ΔLL0ΔT=0.0285 m(50 m)(30 °C)=1.9×10−5 °C−1\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.0285 \text{ m}}{(50 \text{ m}) (30 \text{ °C})} = 1.9 \times 10^{-5} \text{ °C}^{-1}
        b. Identify the metal: Based on the calculated α\alpha value, the metal is likely aluminum.

  10. Concrete sidewalk expansion in Afar region:

    • Given: Slab length L<em>0=3 mL<em>0 = 3 \text{ m}, ΔT=38−25=13 °C\Delta T = 38 - 25 = 13 \text{ °C}, and α</em>concrete≈12×10−6 °C−1\alpha</em>{concrete} \approx 12 \times 10^{-6} \text{ °C}^{-1}.
    • Using the linear expansion formula: ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T
    • ΔL=(12×10−6 °C−1)(3 m)(13 °C)=4.68×10−4 m=0.468 mm\Delta L = (12 \times 10^{-6} \text{ °C}^{-1}) (3 \text{ m}) (13 \text{ °C}) = 4.68 \times 10^{-4} \text{ m} = 0.468 \text{ mm}

  11. Temperature at which bolts touch:

    • Given: Initial gap Δx=5μm=5×10−6 m\Delta x = 5 \mu\text{m} = 5 \times 10^{-6} \text{ m} at 27°C. Assume the bolts are made of steel.
    • The problem requires additional information (e.g., length and material of the bolts). Let's assume the bolts each expand by 5×10−62 m\frac{5 \times 10^{-6}}{2} \text{ m}
    • ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T
      • Need to find the final temperature

  12. Surface area change of an iron drain cover:

    • Given: Initial area A<em>0=0.67 m2A<em>0 = 0.67 \text{ m}^2 at 10 °C, final temperature 105 °C, and β</em>iron=2.2×10−5 K−1\beta</em>{iron} = 2.2 \times 10^{-5} \text{ K}^{-1}.
    • Using the area expansion formula: ΔA=βA0ΔT\Delta A = \beta A_0 \Delta T
    • ΔT=105−10=95 °C\Delta T = 105 - 10 = 95 \text{ °C}
    • ΔA=(2.2×10−5 K−1)(0.67 m2)(95 °C)=1.3927×10−4 m2\Delta A = (2.2 \times 10^{-5} \text{ K}^{-1}) (0.67 \text{ m}^2) (95 \text{ °C}) = 1.3927 \times 10^{-4} \text{ m}^2

  13. Volume change of a copper cube:

    • Given: Side length L=40 cm=0.4 mL = 40 \text{ cm} = 0.4 \text{ m}, initial temperature 20 °C, final temperature 120 °C, and αcopper=1.7×10−5 °C−1\alpha_{copper} = 1.7 \times 10^{-5} \text{ °C}^{-1}.
    • The volume expansion coefficient γ=3α=3(1.7×10−5 °C−1)=5.1×10−5 °C−1\gamma = 3\alpha = 3(1.7 \times 10^{-5} \text{ °C}^{-1}) = 5.1 \times 10^{-5} \text{ °C}^{-1}.
    • Initial volume V0=L3=(0.4 m)3=0.064 m3V_0 = L^3 = (0.4 \text{ m})^3 = 0.064 \text{ m}^3
    • Change in temperature ΔT=120−20=100 °C\Delta T = 120 - 20 = 100 \text{ °C}
    • Using the volume expansion formula: ΔV=γV0ΔT\Delta V = \gamma V_0 \Delta T
    • ΔV=(5.1×10−5 °C−1)(0.064 m3)(100 °C)=3.264×10−4 m3\Delta V = (5.1 \times 10^{-5} \text{ °C}^{-1}) (0.064 \text{ m}^3) (100 \text{ °C}) = 3.264 \times 10^{-4} \text{ m}^3