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

Step-by-step 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 * C * \Delta T`

=
`355 g * 4.18 J/g^(o)C * (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 * \frac{\text{1 mol ice}}{6 kJ} * (18 g)/(1 mol)`

= 111.40 g

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