Question

A platinum resistance thermometer has resistance of 52.5 ohms and 9.75 ohms at 0 degrees celsius and 100 degrees celsius respectively.when the resistance is 8.25 ohms, find the temperature

Answer 1

Answer on Question #45851 – Physics – Other

A platinum resistance thermometer has resistance of 52.5 ohms and 9.75 ohms at 0 degrees celsius and 100 degrees celsius respectively. when the resistance is 8.25 ohms, find the temperature

Solution:

$R_0 = 52.5\Omega$ – initial resistance;

$T_0 = 0{}^\circ C$ – initial temperature;

$R_1 = 9.75\Omega$ – resistance at temperature $T_1 = 100{}^\circ C$

$R_2 = 8.25\Omega$ – resistance at temperature $T_2$

$\alpha$ – temperature coefficient of resistance;

An intuitive approach to temperature dependence leads one to expect a fractional change in resistance which is proportional to the temperature change:

$$R_1 = R_0 (1 + \alpha (T_1 - T_0))$$ $$R_1 = R_0 + R_0 \alpha (T_1 - T_0)$$ $$\alpha = \frac{R_1 - R_0}{R_0 (T_1 - T_0)}$$

Formula for the resistance at temperature $T_2$.

$$R_2 = R_0 \left(1 + \alpha (T_2 - T_0)\right)$$ $$R_2 = R_0 + R_0 \alpha T_2 - R_0 \alpha T_0$$ $$T_2 = \frac{R_2 - R_0 + R_0 \alpha T_0}{R_0 \alpha}$$

(1) in (2):

$$T_2 = \frac{R_2 - R_0 + R_0 T_0 \cdot \frac{R_1 - R_0}{R_0 (T_1 - T_0)}}{R_0 \cdot \frac{R_1 - R_0}{R_0 (T_1 - T_0)}} = \frac{R_0 (T_1 - T_0) (R_2 - R_0) + R_0 T_0 (R_1 - R_0)}{R_0 (R_1 - R_0)} = \frac{52.5\Omega (100{}^\circ C - 0{}^\circ C) (8.25\Omega - 52.5\Omega) + 52.5\Omega \cdot 0{}^\circ C \cdot (9.75\Omega - 52.5\Omega)}{52.5\Omega (9.75\Omega - 52.5\Omega)} = 103.5{}^\circ C$$

Answer: $103.5{}^\circ C$

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