Question

The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD for aluminum, given that the specific heat is 4.6J/kgK at 15K. (A is the constant that related temperature to Cv). (Remember that Cv is not specific heat).

Answer 1

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

$C_{v} =AT^{3}$

Substituting in the expression of the question:

$C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3} } )T^{3}$

$\theta _{D} =(\frac{12\pi ^{4}RT^{3} }{5C_{v} } ) ^{1/3}$

Here

$C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K$

Replacing:

$\theta _{D} =(\frac{12\pi ^{4}*8.31*15^{3} }{5*0.1241} )^{1/3} =375.2361K$