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Question Number 74358 by jagannath19 last updated on 23/Nov/19

Commented by mr W last updated on 23/Nov/19

(4) is correct  length of copper rod at temperature Δθ:  L_0 +L_0 α_S Δθ  length of copper rod at room temperatur:  L_0 +L_0 α_S Δθ−L_0 α_C Δθ  =L_0 [1+(α_S −α_C )Δθ]

$$\left(\mathrm{4}\right)\:{is}\:{correct} \\ $$$${length}\:{of}\:{copper}\:{rod}\:{at}\:{temperature}\:\Delta\theta: \\ $$$${L}_{\mathrm{0}} +{L}_{\mathrm{0}} \alpha_{{S}} \Delta\theta \\ $$$${length}\:{of}\:{copper}\:{rod}\:{at}\:{room}\:{temperatur}: \\ $$$${L}_{\mathrm{0}} +{L}_{\mathrm{0}} \alpha_{{S}} \Delta\theta−{L}_{\mathrm{0}} \alpha_{{C}} \Delta\theta \\ $$$$={L}_{\mathrm{0}} \left[\mathrm{1}+\left(\alpha_{{S}} −\alpha_{{C}} \right)\Delta\theta\right] \\ $$

Commented by jagannath19 last updated on 23/Nov/19

sir i didnt understand please provide detailed soln sir

$${sir}\:{i}\:{didnt}\:{understand}\:{please}\:{provide}\:{detailed}\:{soln}\:{sir} \\ $$

Commented by mr W last updated on 23/Nov/19

when the rod is measured with the  steel tape, the tape shows a length L_0 .  but L_0  is only the scale on the tape.  and the scale shows only the true  length of the tape at room temperature.   the true length of the tape at Δθ is  in fact L_0 +L_0 α_S Δθ. this is also the   measured true length of the rod at Δθ,  the length of the rod at room temperature  is thus L_0 +L_0 α_S Δθ−L_0 α_C Δθ.

$${when}\:{the}\:{rod}\:{is}\:{measured}\:{with}\:{the} \\ $$$${steel}\:{tape},\:{the}\:{tape}\:{shows}\:{a}\:{length}\:{L}_{\mathrm{0}} . \\ $$$${but}\:{L}_{\mathrm{0}} \:{is}\:{only}\:{the}\:{scale}\:{on}\:{the}\:{tape}. \\ $$$${and}\:{the}\:{scale}\:{shows}\:{only}\:{the}\:{true} \\ $$$${length}\:{of}\:{the}\:{tape}\:{at}\:{room}\:{temperature}.\: \\ $$$${the}\:{true}\:{length}\:{of}\:{the}\:{tape}\:{at}\:\Delta\theta\:{is} \\ $$$${in}\:{fact}\:{L}_{\mathrm{0}} +{L}_{\mathrm{0}} \alpha_{{S}} \Delta\theta.\:{this}\:{is}\:{also}\:{the}\: \\ $$$${measured}\:{true}\:{length}\:{of}\:{the}\:{rod}\:{at}\:\Delta\theta, \\ $$$${the}\:{length}\:{of}\:{the}\:{rod}\:{at}\:{room}\:{temperature} \\ $$$${is}\:{thus}\:{L}_{\mathrm{0}} +{L}_{\mathrm{0}} \alpha_{{S}} \Delta\theta−{L}_{\mathrm{0}} \alpha_{{C}} \Delta\theta. \\ $$

Commented by mr W last updated on 23/Nov/19

Commented by jagannath19 last updated on 23/Nov/19

sir you have an amazing sense of knowledge... thank you so much sir...

$${sir}\:{you}\:{have}\:{an}\:{amazing}\:{sense}\:{of}\:{knowledge}...\:{thank}\:{you}\:{so}\:{much}\:{sir}... \\ $$

Commented by mr W last updated on 23/Nov/19

fine, if i could help you!

$${fine},\:{if}\:{i}\:{could}\:{help}\:{you}! \\ $$

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