The electrical resistance in ohms of a certain thermometer varies… - Vidyadip

Question

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
$\text{R}=\text{R}_0[1+\alpha(\text{T}-\text{T}_0)]$
The resistance is 101.6W at the triple-point of water 273.16K, and 165.5W at the normal melting point of lead (600.5K). What is the temperature when the resistance is 123.4W?

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Answer

It is given that:
$\text{R}=\text{R}_0[1+\alpha(\text{T}-\text{T}_0)]\ ...(1)$
where,
R₀ and T₀ are the initial resistance and temperature respectively.
$\alpha$ is a constant
At the triple point of water, T₀ = 273.15K
Resistance of lead, $\text{R}_0 = 101.6\Omega$
At normal melting point of lead, T = 600.5K
Resistance of lead, $\text{R}=165.5\Omega$
Substituting these values in equation (i), we get:
$\text{R}=\text{R}_0[1+\alpha(\text{T}-\text{T}_0)]$
$165.5=101.6[1+\alpha(600.5-273.15)]$
$1.629=1+\alpha(327.35)$
$\therefore\alpha=\frac{0.629}{327.35}=1.92\times10^{-3}\text{K}^{-1}$
For resistance, $\text{R}_1=123.4\Omega$ The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
$\text{R}=\text{R}_0[1+\alpha(\text{T}-\text{T}_0)]$
where,
T is the temperature when the resistance of lead is $123.4\Omega$
123.4 = 101.6 [1 + 1.92 × 10^-3( T – 273.15)]
Solving for T, we get
T = 384.61K.

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