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Thermal Properties of Matter

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Question : 3 of 22
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The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R=R0[1+α(T−T0)]R = R_0 [1 + \alpha (T - T_0)] 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=273.16KR_0 = 101.6 \Omega ; T_0 = 273.16 \mathrm{K}
Case (i) R1=165.5Ω;T1=600.5KR_1 = 165.5 \Omega ; T_1 = 600.5 \mathrm{K}
Case (ii) R2=123.4Ω;T2=?R_2 = 123.4 \Omega ; T_2 = ?
Using the relation R=R0[1+α(T−T0)]R = R_0 [1 + \alpha (T - T_0)]
Case (i) 165.5=101.6[1+α(600.5−273.16)]165.5 = 101.6[1 + \alpha (600.5 - 273.16)] α=165.5−101.6101.6×(600.5−273.16)\alpha = \frac{165.5 - 101.6}{101.6 \times (600.5 - 273.16)} =63.9101.6×327.34= \frac{63.9}{101.6 \times 327.34}
Case (ii) 123.4=101.6[1+α(T2−273.16)]123.4 = 101.6 [1 + \alpha (T_2 - 273.16)]
or
123.4=101.6[1+63.9101.6×327.34(T2−273.16)]123.4 = 101.6 \left[1 + \frac{63.9}{101.6 \times 327.34} (T_2 - 273.16)\right]
or T2=(123.4−101.6)×327.3463.9+273.16T_2 = \frac{(123.4 - 101.6) \times 327.34}{63.9} + 273.16 =111.67+273.16=384.83K= 111.67 + 273.16 = 384.83 \mathrm{K}
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