Question

A platinum resistance thermometer has a resistance of 50 ohm at 20 degree celcius. when dipped in a liquid the resistance becomes 76.8 ohm . the temperature coefficient of resistance for platinum is alpha=3.92×10^-3/°C . the temperature of liquid is ​

Answer 1

Given:

R1 = 50 Ω

T1 = 20°C

R2 = 76.8 Ω

α = 3.92×$10^{-3$ °C .

To find:

T2 =?

Explanation:

The temperature coefficient of resistance is given by

$\displaystyle R_T = R_0(1 + \alpha \triangle T)$         ...(1)

A platinum resistance thermometer has a resistance of 50Ω at 20°C.

R1 = 50Ω, T1= 20°C

Put these values in eq (1)

50 = $R_0$ {( 1 + (3.92 × $10^{-3$ × 20)}         ...(2)

When a platinum thermometer is dipped in a liquid of resistance 76.8 Ω.

R2 = 76.8 Ω

Put these values in eq (1)

76.8 = $R_0$ {1 + ( 3.92 × $10^{-3$  × T2)}      ...(3)

Divide eq (2) to (3)

$\displaystyle \frac{76.8} {50} = \frac{R_o( 1 + 3.92 \times 10^{-3} \times T2)}{ {R_0( 1 + 3.92 \times 10^{-3} \times 20)}}$

Solving the above equation, we get

T2 = 167°C

Hence the temperature of the liquid is ​167°C.