Answer:
The thermal conductivity [tex]k = 1.4094 W/ m\cdot K[/tex]
Explanation:
From the question we are told that
The depth of the thermocouple from the surface is x = 10 mm = 0.01 m
The temperature is [tex]T_f = 100 ^o C[/tex]
The initial temperature is [tex]T_i = 30 ^o C[/tex]
The temperature of the thermocouple after t = 2 minutes( 2 * 60 = 120 \ seconds) is [tex]T_t = 65 ^o C[/tex]
The density of the material is [tex]\rho = 2200 kg/m^3[/tex]
The specific heat of the solid [tex]c_s = 700 J/kg \cdot K[/tex]
Generally the equation for semi -infinite medium is mathematically as
[tex]\frac{T_s - T }{T_i - T} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]\frac{65 - 100 }{30 - 100} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]0.5 = erf [\frac{0.01}{2 \sqrt{\alpha * 120} } ][/tex]
Here [tex]\alpha[/tex] is a constant with unit [tex]m^2 /s[/tex]
[tex]\frac{0.01}{ 2 (\sqrt{\alpha * 120 } )}[/tex] this is from the Gaussian function table
[tex]0.0 1 = 0.954 * (\sqrt{\alpha * 120 } )[/tex]
=> [tex]\sqrt{\alpha * 120 } = \frac{0.01 }{0.954 }[/tex]
=> [tex]\alpha = 9.1525 *10^{-7} \ m^2 /s[/tex]
Generally the thermal conductivity is mathematically represented as
[tex]k = \alpha * \rho * c_s[/tex]
[tex]k = 9.1525 *10^{-7} * 2200 * 700[/tex]
[tex]k = 1.4094 W/ m\cdot K[/tex]
