Answer to Question #293474 in Civil and Environmental Engineering for Alan Enrico V Tuib

Question

A 200-kg sample of Granite is heated to a temperature of 98°C and elongates by 25mm. If its

temperature before it is heated is 25°C, specific heat is 0.80 J/g-K and its length extends to

250mm, 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= (0.80 J/g-K*225mm)/73

K= 2.4658

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=(250mm-25mm)/(98°C-25°C)

α= 225mm/73°C

α=3.0822