Physics, asked by miracleleonard12, 3 months ago

a platinum resistance thermometer wire has a resistance of 5ohm at 0 degree celsius and 5.50ohm at 100 degree celsius. calculate the temperature of the wire when the resistance is 5.2ohm.

Answers

Answered by itzinnocentbndii

Answer:

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Answered by Thorragnarok57

Explanation:

Here R

Here R 0

Here R 0 =5Ω,

Here R 0 =5Ω, R

Here R 0 =5Ω, R 100

Here R 0 =5Ω, R 100 =5.23Ω,R

Here R 0 =5Ω, R 100 =5.23Ω,R t

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795Ω

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α=

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×t

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t=

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100=

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100= 5.23−5

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100= 5.23−55.795−5

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100= 5.23−55.795−5 ×100

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100= 5.23−55.795−5 ×100=346.65

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100= 5.23−55.795−5 ×100=346.65 ∘

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100= 5.23−55.795−5 ×100=346.65 ∘ C

Here R 0 =5Ω, R 100 =5.23Ω,R t =5.795ΩAs α= R 0 ×tR 100 −R 0 ∴t= R 100 −R 0 R t −R 0 ×100= 5.23−55.795−5 ×100=346.65 ∘ CHence, this is the answer.

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