A 175-kg sample of Copper is heated to a temperature of 116°C and elongates by 35mm. If
its temperature before it is heated is 20°C, specific heat is 0.38 J/g-K and its length extends to
330 mm, calculate the following:
a. Heat absorbed or thermal conductivity
b. Linear expansion coefficient
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= (0.38 J/g-K*330mm)/96
K= 1.30625
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=(330mm-35mm)/(116°C-20°C)
α= 295mm/96°C
α=3.073
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