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Asked by Gaurav Last Modified
Answer As an experienced tutor registered on UrbanPro, I'd be happy to assist you with this problem. UrbanPro is known for connecting students with expert tutors for personalized learning experiences.
For this problem, we're dealing with the relationship between electrical resistance and temperature, which is given by the formula:
R=R0[1+α(T−T0)]R=R0[1+α(T−T0)]
where:
Given that the resistance at the triple point of water (T0=273.16 KT0=273.16K) is R0=101.6 ΩR0=101.6Ω and the resistance at the normal melting point of lead (T=600.5 KT=600.5K) is R=165.5 ΩR=165.5Ω, we can use these values to find the value of αα.
First, let's rearrange the equation to solve for αα:
α=R−R0R0(T−T0)α=R0(T−T0)R−R0
Substituting the given values:
α=165.5−101.6101.6×(600.5−273.16)α=101.6×(600.5−273.16)165.5−101.6
Now, we can find the value of αα.
Once we have αα, we can use it to find the temperature TT when the resistance is R=123.4 ΩR=123.4Ω. Rearranging the equation, we get:
T=T0+R−R0αR0T=T0+αR0R−R0
Substitute the given values into this equation to find the temperature corresponding to the resistance of 123.4 Ω123.4Ω.
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