Question

An insulated Thermos contains 150 g of water at 87.3 ˚C. You put in a 10.2 g ice cube at 0.00 ˚C to form a system of ice + original water. The specific heat of liquid water is 4190 J/kg•K; and the heat of fusion of water is 333 kJ/kg. What is the net entropy change of the system from then until the system reaches the final (equilibrium) temperature?

Answer 1

Step-by-step explanation:

Let us take the equilibrium temperature is T.

Now, the heat absorbed by the ice is equal to the heat lost to the water.

Therefore, the formula will be as follows.

`m_(water) * C_(water) (T - 87.3) + m_(ice) * L_(f) + m_(ice) * C_(water) (T - 0) = 0`

Therefore, putting the given values into the above formula as follows.

`m_(water) * C_(water) (T - 87.3) + m_(ice) * L_(f) + m_(ice) * C_(water) (T - 0) = 0`

`150 g * 4190 J/kg K (T - 87.3) + 0.0102 kg * 333 kJ/kg + m_(ice) * C_(water) (T - 0) = 0`

628500T - 54868050 + 3.3966 + 42.738T = 0

T =
`(54868046.6)/(628542.738)`

=
`87.29^(o)C`

or, = (87.29 + 273.15)K

= 360.44 K

Since, formula for entropy change of ice is as follows.

`\Delta S_(ice) = (m_(ice) * L_(f))/(T_(ice)) + m_(ice) C ln (T)/(T_(melting))`

=

=
`0.0102 * 2381.015`

= 24.286 J/K

Entropy change for water (
`\Delta S_(water)`) =
`m_(ice) * C * ln ((T)/(T_(initial)))`

=
`0.0102 * 4190 * ln (360.44)/(360.45)`

=
`-1.186 * 10^(-3)` J/K

Therefore, calculate the net change in entropy as follows.

`\Delta S = \Delta S_(water) + \Delta S_(ice)`

=
`-1.186 * 10^(-3) + 24.286` J/K

= 24.28481 J/K

Thus, we can conclude that the net entropy change of the system from then until the system reaches the final (equilibrium) temperature is 24.28481 J/K.