Answer to Question #293475 in Civil and Environmental Engineering for Alan Enrico V Tuib

Question

A 150-kg sample of Steel is heated to a temperature of 143°C and elongates by 58 mm. If its

temperature before it is heated is 37°C, specific heat is 0.46 J/g-K and its length extends to

420 mm, calculate the following:

a. Heat absorbed or thermal conductivity

b. Linear expansion coefficient

Answer 1

a. Heat absorbed or thermal conductivity

K = (QL)/(AΔT), Where;

K is the thermal conductivity in W/m.K.

Q is the amount of heat transferred through the material in Joules/second or Watts.

L is the distance between the two isothermal planes.

A is the area of the surface in square meters.

ΔT is the difference in temperature in Kelvin.

K= (0.46 J/g-K*362mm)/106

K= 1.571

b. Linear expansion coefficient

ΔL = αLΔT, Where;

ΔL is the change in length L,

ΔT is the change in temperature, and

α is the coefficient of linear expansion, which varies slightly with temperature.

α=ΔL/ΔT

α=(420mm-58mm)/(143°C-37°C)

α= 362mm/106°C

α=3.4151