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

Step-by-step explanation:

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

Q =
`m * C * \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 * C * \Delta T`

=
`474 g * 4.184 J/g^(o)C * 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.

`(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`.