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A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon resistor is 217.7 Ω. What is the temperature on a spring day when the resistance is 215.1 Ω? Take the temperature coefficient of resistivity for carbon to be α = −5.00×10^−4C^-1

Respuesta :

lublana

Answer:

28 degree C

Explanation:

We are given that

[tex]T_1=4.00^{\circ}[/tex]

[tex]R_1=217.7 \Ohm[/tex]

[tex]R_2=215.1\Ohm[/tex]

[tex]\alpha=-5.00\times 10^{-4}C^{-1}[/tex]

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

[tex]\alpha(T_2-T_1)=\frac{R_2}{R_2}-1[/tex]

Using the formula

[tex]-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1[/tex]

[tex]-5\times 10^{-4}(T_2-4)=0.988-1=-0.012[/tex]

[tex]T_2-4=\frac{0.012}{5\times 10^{-4}}=24[/tex]

[tex]T_2=24+4=28^{\circ}C[/tex]

Hence, the temperature  on a spring day 28 degree C.

asuwafohjames

The Temperature on a spring day is 28°C.

The problem above can be solved using the formula below.

⇒ Formula:

  • R = R'[1+α(T₂-T₁)]............... Equation 1

⇒ Where:

  • R = New resistance
  • R' = Original resistance
  • α = Temperature coefficient of resistivity
  • T₂ = Final Temperature
  • T₁ = Initial Temperature.

From the question,

⇒ Given:

  • R = 215.1 Ω
  • R' = 217.7 Ω
  • T₁ = 4.00 °C
  • T₂ = ?
  • α = -5.0×10⁻⁴ C⁻¹

⇒ Substitute these values into equation 1

  • 215.1 = 217.7[1+(-5×10⁻⁴)(T₂-4)]

⇒ Solve for T₂

  • [1+(-0.0005)(T₂-4)] = 215.1/217.7
  • (-0.0005)(T₂-4) = 0.988-1
  • (T₂-4) = -0.012/-0.0005
  • (T₂-4) = 24
  • T₂ = 24+4
  • T₂ = 28°C

Hence, The Temperature on a spring day is 28°C.

Learn more about Temperature here: https://brainly.com/question/16559442

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