11.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :

R = Ro [1 + α(T – $T_{0}$ )]

The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?

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Payal Gupta | Contributor-Level 10

5 months ago

11.3 It is given that R = $R_{0}$ [1 + α(T – $T_{0}$ )] ………(i)

Where Ro and To are the initial resistance and temperature and R and T are the final resistance and temperature.

At the triple point of water, To = 273.15 K, $R_{0}$ = 101.6 Ω

At normal melting point of lead, T = 600.5 K, R = 165.5 Ω

Substituting these values in equation (i), we get

165.5 = $101.6$ [1 + α(600.5 – $273.15$ )]

$\frac{165.5}{101.6}$ = 1 + 327.35 α, or α = $\frac{0.629}{327.35}$ = 1.92 $\times 10^{- 3}$ $K^{- 1}$

When R = 123.4 Ω, T can be calculated as

123.4 = $101.6$ [1 + 1.92 $\times 10^{- 3} \times$ (T – $273.15$ )]

1.214 =

...more

11.3 It is given that R = $R_{0}$ [1 + α(T – $T_{0}$ )] ………(i)

Where Ro and To are the initial resistance and temperature and R and T are the final resistance and temperature.

At the triple point of water, To = 273.15 K, $R_{0}$ = 101.6 Ω

At normal melting point of lead, T = 600.5 K, R = 165.5 Ω

Substituting these values in equation (i), we get

165.5 = $101.6$ [1 + α(600.5 – $273.15$ )]

$\frac{165.5}{101.6}$ = 1 + 327.35 α, or α = $\frac{0.629}{327.35}$ = 1.92 $\times 10^{- 3}$ $K^{- 1}$

When R = 123.4 Ω, T can be calculated as

123.4 = $101.6$ [1 + 1.92 $\times 10^{- 3} \times$ (T – $273.15$ )]

1.214 = 1 + T1.92 $\times 10^{- 3}$ - 1.92 $\times 10^{- 3} \times$ 273.15

T = 384.6 K

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11.3 It is given that R = <math><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math> [1 + α(T – <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math> )] ………(i)Where Ro and To are the initial resistance and temperature and R and T are the final resistance and temperature.At the triple point of water, To = 273.15 K, <math><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math> = 101.6 ΩAt normal melting point of lead, T = 600.5 K, R = 165.5 ΩSubstituting these values in equation (i), we get165.5 = <math><mn>101.6</mn></math> [1 + α(600.5 – <math><mn>273.15</mn></math> )]<math><mfrac><mrow><mrow><mn>165.5</mn></mrow></mrow><mrow><mrow><mn>101.6</mn></mrow></mrow></mfrac></math> = 1 + 327.35 α, or α = <math><mfrac><mrow><mrow><mn>0.629</mn></mrow></mrow><mrow><mrow><mn>327.35</mn></mrow></mrow></mfrac></math> = 1.92 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> <math><msup><mrow><mrow><mi>K</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math>When R = 123.4 Ω, T can be calculated as123.4 = <math><mn>101.6</mn></math> [1 + 1.92 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup><mo>×</mo></math> (T – <math><mn>273.15</mn></math> )]1.214 = 1 + T1.92 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> - 1.92 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup><mo>×</mo></math> 273.15T = 384.6 K

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