Test Index

Thermal Properties of Matter

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Question : 3
Total: 22
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R=R0[1+α(T–T0)] 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?
Solution:  
Here, R0=101.6Ω;T0=273.16K
Case (i) R1=165.5Ω;T1=600.5K
Case (ii) R2=123.4Ω;T2=?
Using the relation R=R0[1+α(T−T0)]
Case (i) 165.5=101.6[1+α(600.5−273.16)] α=
165.5−101.6
101.6×(600.5−273.16)
 =
63.9
101.6×327.34

Case (ii) 123.4=101.6[1+α(T2−273.16)]
or
123.4=101.6[1+
63.9
101.6×327.34
(T2−273.16)]
or T2=
(123.4−101.6)×327.34
63.9
+273.16 =111.67+273.16=384.83K
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