Question

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = Ro [1 + α (T – To)]
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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Answer 1

Explanation:

It is given that:

R = R0 [1 + α (T – T0)] … (i)

where,

R0 and T0 are the initial resistance and temperature respectively

R and T are the final resistance and temperature respectively

α is a constant

At the triple point of water, T0 = 273.15 K

Resistance of lead, R0 = 101.6 Ω

At normal melting point of lead, T = 600.5 K

Resistance of lead, R = 165.5 Ω

Substituting these values in equation (i), we get:

R = Ro [1 + α (T – To)]

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

1.629 = 1 + α (327.35)

∴ α = 0.629 / 327.35 = 1.92 × 10-3 K-1

For resistance, R1 = 123.4 Ω

R1 = R0 [1 + α (T – T0)]

where,

T is the temperature when the resistance of lead is 123.4 Ω

123.4 = 101.6 [ 1 + 1.92 × 10-3( T - 273.15) ]

Solving for T, we get

T = 384.61 K.

Answer 2

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It is given that:

It is given that:R = R0 [1 + α (T – T0)] … (i)

It is given that:R = R0 [1 + α (T – T0)] … (i)where,

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectively

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectively

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constant

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 K

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 KResistance of lead, R0 = 101.6 Ω

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 KResistance of lead, R0 = 101.6 ΩAt normal melting point of lead, T = 600.5 K

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 KResistance of lead, R0 = 101.6 ΩAt normal melting point of lead, T = 600.5 KResistance of lead, R = 165.5 Ω

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 KResistance of lead, R0 = 101.6 ΩAt normal melting point of lead, T = 600.5 KResistance of lead, R = 165.5 ΩSubstituting these values in equation (i), we get:

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 KResistance of lead, R0 = 101.6 ΩAt normal melting point of lead, T = 600.5 KResistance of lead, R = 165.5 ΩSubstituting these values in equation (i), we get:R = Ro [1 + α (T – To)]

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 KResistance of lead, R0 = 101.6 ΩAt normal melting point of lead, T = 600.5 KResistance of lead, R = 165.5 ΩSubstituting these values in equation (i), we get:R = Ro [1 + α (T – To)]165.5 = 101.6 [ 1 + α(600.5 – 273.15) ]

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 KResistance of lead, R0 = 101.6 ΩAt normal melting point of lead, T = 600.5 KResistance of lead, R = 165.5 ΩSubstituting these values in equation (i), we get:R = Ro [1 + α (T – To)]165.5 = 101.6 [ 1 + α(600.5 – 273.15) ]1.629 = 1 + α (327.35)

It is given that:R = R0 [1 + α (T – T0)] … (i)where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectivelyα is a constantAt the triple point of water, T0 = 273.15 KResistance of lead, R0 = 101.6 ΩAt normal melting point of lead, T = 600.5 KResistance of lead, R = 165.5 ΩSubstituting these values in equation (i), we get:R = Ro [1 + α (T – To)]165.5 = 101.6 [ 1 + α(600.5 – 273.15) ]1.629 = 1 + α (327.35)∴ α = 0.629 / 327.35 = 1.92 × 10-³ K-1

K-1For resistance, R1 = 123.4 Ω

K-1For resistance, R1 = 123.4 ΩR1= R0 [1 + α (T – T0)]

K-1For resistance, R1 = 123.4 ΩR1= R0 [1 + α (T – T0)]where,

K-1For resistance, R1 = 123.4 ΩR1= R0 [1 + α (T – T0)]where,T is the temperature when the resistance of lead is 123.4 Ω

K-1For resistance, R1 = 123.4 ΩR1= R0 [1 + α (T – T0)]where,T is the temperature when the resistance of lead is 123.4 Ω123.4 = 101.6 [ 1 + 1.92 × 10-³( T – 273.15) ]

( T – 273.15) ]Solving for T, we get

( T – 273.15) ]Solving for T, we getT = 384.61 K.

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