How to Calculate the Average Molar Heat Capacity of NaCl at Low Temperatures?

[Je m'appelle]

Homework Statement

The molar heat capacity at a constant volume of a solid at low temperatures T << Td, where Td is the Debye temperature is given by:

C_v = 464(\frac{T}{T_d})^3

Consider Td = 281 K for the NaCl.

(a) Calculate the average molar heat capacity \bar{C_v} of the NaCl between the temperatures of Ti = 10 K and Tf = 20 K.

(b) Calculate the amount of heat necessary to raise the temperature of 1,000g of NaCl from 10 K to 20 K.

Homework Equations

C_v = 464(\frac{T}{T_d})^3

C_v = (\frac{\delta Q}{dT})_v

The Attempt at a Solution

(a) So I found the heat capacities Ci and Cf at Ti and Tf, respectively, and then in order to find the average heat capacity I summed them and divided by 2.

C_i = 464(\frac{10}{281})^3

C_i = 0,021

C_f = 464(\frac{20}{281})^3

C_f = 0,167

\bar{C_v} = \frac{C_i + C_f}{2}

\bar{C_v} = 0,94

Is this correct?

(b)

(\frac{\delta Q}{dT})_v = C_v

\delta Q = \bar{C_v} dT

\delta Q = (0,94)(10) = 9,4

Is this also correct?

Thanks.

 

[zachzach]

I don't think the first part is correct since:

<br /> f_{av} = \frac{1}{b-a} \int_{a}^{b} f(x) dx <br /> <br />

 

[Je m'appelle]

I see my mistake now, thank you very much kind sir.