Question

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.

Answer 1

Explanation:

The given data is as follows.

      Mass of ice dropped = 325 g

      Initial temperature = $30^{o}C$ = (30 + 273) K = 303 K

     Final temperature = $0.0^{o}C$ = (0 + 273) K = 273 K

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

             $m_{ice} = \frac{500 ml \times 1.0 g/mol}{1 ml}$

                        = 500 g

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

                 Q = $m \times C_{p} \times dT$

                      = $\frac{500 g}{18 g/mol} \times 75.3 J/K/mol \times (303 - 273)K$

                      = 62750 J

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

                       Q = m L

              = $\frac{325 g}{18 g/mol} \times 6 \times 10^{3}$

             = 108300 J

Therefore, we require $\frac{325 g}{18 g/mol} J$ heat but we have 40774.95 J.

So,     $mass_{ice} = \frac{m \times C_{p} \times (T_{f} - T_{i})}{\Delta H_{f}}$

                      = $\frac{62750 J}{333}$

                      = 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.