Question #199590

The ratio of linear expansivity of copper to that of iron Is approximately 1:5. A specimen of iron and specimen of copper expand by the same amount per unit rise in temperature. What is the ratio of the length of iron to that of copper ?


1
Expert's answer
2021-05-27T17:47:48-0400

The expansion per unit rise in temperature is given as follows:


ΔLΔT=αL\dfrac{\Delta L}{\Delta T} = \alpha L

where ΔL\Delta L is the increase in length, ΔT\Delta T is the rise in temperature, α\alpha is the expansivity, and LL is the initial length.

According to the text of the problem, (ΔLΔT)cooper=(ΔLΔT)iron\left( \dfrac{\Delta L}{\Delta T} \right)_{cooper } = \left( \dfrac{\Delta L}{\Delta T} \right)_{iron } and αcooperαiron=15\dfrac{\alpha_{cooper}}{\alpha_{iron}} = \dfrac15. Then the ration of the lengths is:


Lcooper=1αcooper(ΔLΔT)cooperLiron=1αiron(ΔLΔT)ironLironLcooper=1αiron(ΔLΔT)iron1αcooper(ΔLΔT)cooper=αcooperαiron=15L_{cooper} = \dfrac{1}{\alpha_{cooper}} \left( \dfrac{\Delta L}{\Delta T} \right)_{cooper } \\ L_{iron} = \dfrac{1}{\alpha_{iron}} \left( \dfrac{\Delta L}{\Delta T} \right)_{iron }\\ \dfrac{L_{iron} }{L_{cooper} } = \dfrac{\dfrac{1}{\alpha_{iron}} \left( \dfrac{\Delta L}{\Delta T} \right)_{iron }}{ \dfrac{1}{\alpha_{cooper}} \left( \dfrac{\Delta L}{\Delta T} \right)_{cooper }} = \dfrac{\alpha_{cooper}}{\alpha_{iron}} = \dfrac15

Answer. 1:5.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS