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

The Debye temperature for aluminum is 375.2361 K

Step-by-step 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) =((12\pi ^(4)R )/(5\theta _(D)^(3) ) )T^(3)`

`\theta _(D) =((12\pi ^(4)RT^(3) )/(5C_(v) ) ) ^(1/3)`

Here

`C_(v) =4.6(J)/(kg-K) *(1kg)/(1000g) *(26.98g)/(1mol) =0.1241J/mol-K`

Replacing:

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