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never [62]
2 years ago
5

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 f

or 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).
Engineering
1 answer:
kirill115 [55]2 years ago
3 0

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

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