Question

A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after the temperature change? A. 0.0000189 meter B. 0.00002363 meter C. 0.00023635 meter D. 0.124764 meter

Answer 1

$\Delta L= \alpha L_0 (T_f-T_i)$

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D

Answer 2

ΔL=Lo*α*ΔT
We know that α=2,0*10⁻⁵
ΔL=0,125m*2,0*10⁻⁵*100°C
ΔL=0,00025m
Now the new lengh would be:
0,125m+0,00025=0,12525 ≈(d)