A wire of length 100.0m at 30o has linear expansive of 2x10-5K-1 Calculate the length of the wire at a temperature of - 10oC
100.08m
100.04m
99.96m
99.92m
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Discussions (28)
99.92 is very correct the problem is that my school didnt use the appropriate formula
and it wasnt broken down for you guys to understand hence
solution 
using the formular L2=L1[1+linear expansivity x temperature rise]
we have that
L2=100[1+0.00002x(-10-30)]
L2=100[1+0.00002x(-40)]
L2=100[1+(-0.0008)]
L2=100 x 0.9992
=99.92
Nope... Thats from a wrong source....
L.e = l2-l1/l1(change of temp)
2 x 10-5 = l2-100 /100(30-(-10)
2 x 10-5 x 4000 = l2-100
8 x 10-2= l2-100
0.08 = l2 -100
l2 = 100 + 0.08
l2= 100.08
Am i making sense....
Make i ma
At first I calculated it n I got 100.08m.... I knew i made some errors so I calculated it again and I got 99.92m...so, my school is correct. Guys solve it again, the answer is 99.92m.
why wasn't the temperature converted to kelvin, where as the linear expensively is in K^-1
don't think the solution is correct formula is suppose to be
L2 = L1(1 + ¥∆temp.)
¥ = linear expansivity
∆temp. = change in temperature
L1=100 m
L2=?
T1=30°c
T2=-10°c
linear expansivity ¶= 0.00002
Note: expansivity increases with Temperature hence your answer should be less than 100 m since temperature dropped from 30°c to -10°c
¶= L2-L1/L1∆T or ∆L/L1∆T
∆L= ¶ x L1 x ∆T
∆L= 0.00002 x 100 x (-10-30)
∆L= 0.00002 x100 x (-40)
∆L= -0.08 ( negative sign denoting the reduction in length with decrease in temperature)
∆L= L2 - L1
L2= ∆L + L1
L2= - 0.08 + 100
L2=99.92 m 
