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.

Explanation:

The given data is as follows.

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

      Initial temp. of block 1, = (100 + 273) K = 373 K  

      Initial temp. of block 2, = (0 + 273) K = 273 K

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

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

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

                         =

                         = 1243550 J

or,                     = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

            =

            =

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

           =

           =

           = 647.49 J/K

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

                       = -554.12 J/K + 647.49 J/K

                       = 93.37 J/K

Thus, we can conclude that for the given reaction is 1243.5 kJ and is 93.37 J/K.

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