You make some iced tea by dropping 325 grams of ice into 500.0 ml of warm tea in an insulated pitcher. if the tea is initially at 30.0°c and the ice cubes are initially at 0.0°c, how many grams of ice will still be present when the contents of the pitcher reach a final temperature? the tea is mostly water, so assume that it has the same density (1.0 g/ml), molar mass, heat capacity (75.3 j/k/mol), and heat of fusion (6.01 kj/mol) as pure water. the heat capacity of ice is 37.7 j/k/mol.

Explanation:

The given data is as follows.

      Mass of ice dropped = 325 g

      Initial temperature = = (30 + 273) K = 303 K

     Final temperature = = (0 + 273) K = 273 K

Now, using density of water calculate the mass of ice as follows.

                        = 500 g

As the relation between heat energy, specific heat and change in temperature is as follows.

                 Q =

                      =

                      = 62750 J

Also, relation between heat energy and latent heat of fusion is as follows.

                       Q = m L

              =

             = 108300 J

Therefore, we require heat but we have 40774.95 J.

So,    

                      =

                      = 188.4 g

Hence, the mass of ice = 325 g - 188.4 g

                                        = 137 g

Therefore, we can conclude that 137 g of ice will still be present when the contents of the pitcher reach a final temperature.

the objects have the same charge.