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Home » The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: The resistance is at the triple-point of water , and at the normal melting point of lead (600.5 K). What is the temperature when the resistance is ?
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: \mathbf{R}=\mathbf{R}_{\mathrm{o}}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)\right] The resistance is 101.6 \Omega at the triple-point of water 273.16 \mathrm{~K}, and 165.5 \Omega at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 \Omega ?
CBSE | Class 11 | NCERT | Thermal Properties of Matter
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: \mathbf{R}=\mathbf{R}_{\mathrm{o}}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)\right] The resistance is 101.6 \Omega at the triple-point of water 273.16 \mathrm{~K}, and 165.5 \Omega at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 \Omega ?

Solution:

Given information in the question,

Triple point temperature, T_{0}=273.16 \mathrm{~K}

Resistance at the triple point, R_{0}=101.6 \Omega

Normal melting point of lead, T_{1}=600.5 \mathrm{~K}

Resistance at normal melting point, R_{1}=165.5 \Omega

According to approximate law, that states isolated physical system’s observable property does not change over time.

\mathrm{R}_{1}=\mathrm{R}_{0}\left[1+\alpha\left(\mathrm{T}_{1}-\mathrm{T}_{0}\right)\right]

165.5=101.6[1+\alpha(600.5-273.16)]

165.5=101.6[1+\alpha(327.34)]

165.5=101.6+\alpha(101.6)(327.34)

165.5=101.6+\alpha(101.6 \times 327.34)

\alpha=(165.5-101.6) /(101.6 \times 327.34)

\alpha=63.9 /(101.6 \times 327.34)

Now when resistance is 123.4 \Omega then temperature T_{2} is:

R_{2}=R_{0}\left[1+\alpha\left(T_{2}-T_{0}\right)\right]

123.4=101.6\left[1+\alpha\left(T_{2}-273.16\right)\right]

123.4=101.6\left[1+(63.9 /(101.6 \times 327.34))\left(T_{2}-273.16\right)\right]

123.4=101.6+(63.9 / 327.34)\left(T_{2}-273.16\right)

123.4=101.6+(0.195)\left(T_{2}\right)-(0.195)(273.16)

123.4=101.6+(0.195)\left(T_{2}\right)-53.32

T_{2}=(123.4-101.6+53.32) / 0.195=75.12 / 0.195=385.23 K

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