A 0.5 m rod of some material elongates 0.1 mm on heating from 20 to 112°C. Determine the value of the linear coefficient of thermal expansion [in (°C)-1] for this material.

linear coefficient of thermal expansion of the rod =2.17 x10^-6(°C)^-1

Explanation:

The Linear thermal expansion is calculated as

ΔL = αLΔT,

where ΔL = change in length L,

ΔT = the change in temperature,

α is the coefficient of linear expansion

Change in length ΔL= 0.1 mm =0.1/1000= 0.0001m

change in temperature,ΔT= Final - Initial temperature

= (112- 20)°C=92°C

Solving, we have that

ΔL = αLΔT

α=ΔL/LΔT

= 0.0001 / 0.5 x 92

=0.0001/46

=2.17 x10^-6 / °C

The  linear coefficient of thermal expansion of the rod, α =2.17 x10^-6(°C)^-1

answer: report them

explanation: