Question

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in an isolated container. The specific heat capacity of copper is 0.385 J K−1 g−1 and may be assumed constant over the temperature range involved.

Answer 1

Explanation:

The given data is as follows.

        m = 10.0 kg = 10,000 g    (as 1 kg = 1000 g)

      Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K  

      Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

            $mC \Delta T = mC \times \Delta T$

  $10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

                  $T_{f} - 100^{o}C = 0^{o}C - T_{f}$    

                   $2T_{f} = 100^{o}C$

                          $T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.              

Also, we know that the relation between enthalpy and temperature change is as follows.

             $\Delta H = mC \Delta T$

                         = $10000 g \times 0.385 J/K g \times 323 K$

                         = 1243550 J

or,                     = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

     $\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

            = $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

            = $10000 g \times 0.385 J/K g \times -0.143$

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

   $\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

           = $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

           = $10000 g \times 0.385 J/K g \times 0.168$

           = 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

            $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

                       = -554.12 J/K + 647.49 J/K

                       = 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.