1 Calculate the heat energy released when 103 g of liquid me

Solution

1.dq = m s dT + mL

    = 10.3 * 28* -13.68 + 10.3 * 2.29kJ/mol

    = -3.945312 kJ + 23.587

= 19.641688 kJ /mol is the energy released in the case of mercury.

2.dq = msdT for (s------>l phase) +mL +msdT (for l--->g phase)

       = 55 *37.7 * 10 + 55* 6.01*1000 +55 *75.3 *119

       =20735 + 330550 + 492838.5 J/mol

       = 20.735 + 330.550 + 492.8385 kJ/mol

      = 844.1235 kJ/mol

3) the required equation for this will be

         - ( 2*20 * 16 * 37.7 + 2*20* 6.01*1000) = 205 * 75.3 * (T2 -25)

===> - (24128 +240400) = 15436.5 (T2-25)

====>(T2-25) = - 264528 / 15436.5

                     = -17.1365

====> T2 = - 17.1365 + 25

               = 7.8635 degree C