Question

Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and the other at Tc, are brought into contact and allowed to reach equilibrium. Evaluate the change for the two blocks of copper, each of mass 500 grams with Cpcm= 24.4 J KT-1 mol-1, taking Th = 500 K and Tc = 250 K.

Answer 1

Explanation:

Relation between entropy change and specific heat is as follows.

            $\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

The given data is as follows.

     mass = 500 g,         $C_{p}$ = 24.4 J/mol K

     $T_{h}$ = 500 K,          $T_{c}$ = 250 K               

   Mass number of copper = 63.54 g /mol

Number of moles = $\frac{mass}{/text{\molar mass}}$

                                 = $\frac{500}{63.54}$

                                 = 7.86 moles

Now, equating the entropy change for both the substances as follows.

     $7.86 \times 24.4 \times [T_{f} - 250]$ = $7.86 \times 24.4 \times [500 -T_{f}]$

       $T_{f} - 250 = 500 - T_{f}$

          $2T_{f}$ = 750

So,       $T_{f}$ = $375^{o}C$

• For the metal block A,  change in entropy is as follows.

         $\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

              = $24.4 log [\frac{375}{500}]$

              = -3.04 J/ K mol

• For the block B,  change in entropy is as follows.

         $\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

                  = $24.4 log [\frac{375}{250}]$

                  = 4.296  J/Kmol

And, total entropy change will be as follows.

                       = 4.296 + (-3.04)

                      = 1.256 J/Kmol

Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol  and change in entropy of block B is 4.296  J/Kmol.