Question

Use the standard enthaplies of formation to calculate the standard change in enthaply for the melting of ice. (-291.8 kj/mol for H20 (s). use this value to calculate the mass of ice required to cool 355 mL of a beverage from the room temperature (25 degree celsius) to 0 degree celsius. Assume that the specfic heat capacity and the density of the beverage are the same as those of water.

Answer 1

Explanation:

The given reaction is as follows.

          $H_{2}O(s) \rightleftharpoons H_{2}O(l)$

Hence, using standard values heat of fusion will be calculated as follows.

     $\Delta H_{fus} = \Delta H^{o}_{f}(H_{2}O(l)) - \Delta H^{o}_{f}(H_{2}O(s))$

                  = -285.8 kJ/mol - (-291.8 kJ/mol)

                  = 6 kJ/mol

Therefore, heat released from beverage will be calculated as follows.

           q = $m \times C \times \Delta T$

               = $355 g \times 4.18 J/g^{o}C \times (0 - 25)^{o}C$

               = -37133 J

or,           = -37.133 kJ         (as 1 J = 0.001 kJ)

Now, heat gained by ice = - heat released by the beverage

                                         = - (-37.133 kJ)

                                         = 37.133 kJ

Therefore, calculate the mass of ice as follows.

          Mass of ice = $37.133 kJ \times \frac{\text{1 mol ice}}{6 kJ} \times \frac{18 g}{1 mol}$

                              = 111.40 g

Thus, we can conclude that mass of ice is 111.40 g.