For some material, the heat capacity at constant volume Cv at 34 K is 0.81 J/mol-K, and the Debye temperature is 306 K. Estimate the heat capacity (in J/mol-K) (a) at 44 K, and (b) at 477 K.

a. Heat Capacity = 1.756J/mol-K

b. Heat Capacity = 24.942J/mol-k

Explanation:

Given

Constant volume Cv = 0.81J/mol-k

T1 = 34K

Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K

First, The value of the temperature-independent constant.

Using Cv = AT³

Make A the subject of formula

A = Cv/T³

Substitute each values

A = 0.81/34³

A = 0.000020608589456543

A = 2.061 * 10^-5J/mol-k

The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature

Cv = AT³

So, The heat capacity when T = 44k is then calculated as

Cv = 2.061 * 10^-5 * 44³

Cv = 1.755522084266232

Cv = 1.756J/mol-K

(b) at 477 K.

Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:

Cv = 3R

Where R = universal gas constant

R = 8.314J/mol-k

Cv = 3 * 8.314

Cv = 24.942J/mol-k

answer:

yes

explanation:

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