Question

Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. The length l of the bar obeys the following relation: l=1.0000+2.4×10−5T, where T is the number of degrees Celsius above room temperature. What is the change of the bar's length if the temperature is raised to 11.3 ∘C above room temperature?

Express your answer in meters to two significant figures.

Answer 1

the length of the bar is 31.5m,
keeping in mind the relation that is obeyed:
l=1.000+2.4×10-5T
reverse the equation to avoid confusion
T5-10×2.4+1.0000=l
now follow P.E.M.D.A.S
because there are no perenthesis or exponents you can eliminate the P and E.
start by multiplying 11.3 by 5
(11.3×5=56.5)
now the equation will read
56.5-10×2.4+1.0000
then multiply 10 by 2.4
(10×2.4=24)
equation reads
56.5-24+1.0000
now that we have the solved all of the multiplication we can move on with p.e.m.D.A.S.
since there is no division to solve we can start the addition
the equation should look like this:
56.5-24+1.0000
now add 24 to 1.0000
(24+1.0000=25)
equation after findin the sum of the addition solved is:
56.5-25=l
finally subtract 25 from 56.5
(56.5-25=31.5)
31.5=l

Answer 2

Answer:

0.00027m

Explanation:

Given the initial length of aluminum bar Lo = 1.0000m

After heating the aluminum bar under a temperate above room temperature which is 11.3°C, its new length is modelled by the equation:

Lf =1.0000+2.4×10^−5T, where T is the number of degrees Celsius above room temperature.

If T = 11.3°C

Lf = 1.0000+(2.4×10^-5)(11.3)

Lf = 1.0000+27.12×10^-5

Lf = 1.0000+0.0002712

Lf = 1.0002712

The change in bar length = Final length Lf - initial length Lo

Change in bar length = 1.0002712-1.0000

Change in bar length = 0.0002712m

= 0.00027m (to 2s.f)