Question

You order a 16 oz glass of tea (where the mass of water is 474 grams) from a local restaurant. The tea is freshly brewed and has an initial temperature of 28.29 °C. You add ice to cool it. If the heat of fusion of ice is 6.020 kJ/mol and each ice cube contains exactly 1 mol of water, how many ice cubes are necessary to cool the tea to 0.41 °C? The specific heat of the "tea" is 4.184 J/g*C.

Answer 1

Explanation:

It is known that relation between heat energy, specific heat and temperature change is as follows.

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

where,    q = heat energy

              m = mass

              C = specific heat

         $\Delta T$ = change in temperature = $(28.29 - 0.41)^{o}C$ = $27.88^{o}C$

Therefore, calculate the heat energy as follows.

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

                   = $474 g \times 4.184 J/g^{o}C \times 27.88^{o}C$

                  = 55292.06 J

or,              = 55.3 kJ        (as 1 kJ = 1000 J)

One ice cube is equal to 1 mole.

Hence, 6.020 kJ/mol of energy absorbed by 1 ice cube. And, number of ice cubes absorbing 55.3 kJ of energy will be calculated as follows.

               $\frac{55.3 kJ}{6.020 kJ}$

                 = 9

Thus, we can conclude that 9  ice cubes are necessary to cool the tea to $0.41^{o}C$.