Question

What is the entropy change when 5.20 g of ice at 0.0°C are added to 250 mL of water in an insulated thermos at 30.0°C?

answer in (J/K)

Answer 1

The enthalpy of fusion of ice = 6.01kJ/mol and specific heat of water = 4.18J /g

The changes involved in the process are

1) H2O (s) -----------> H2O(l) -------------> H2O(l)

5.2 g at 0 C q1 at 0C q2 at TK

2) H2O (l) -----------------> H2O(l)

250mL at 303 K q3 at TK

The heat absorbed by ice to melet to Tk =- heat given by water at 30C

that is q1 + q2 = -q3

Now q1 = (mass /molar mass)x enthalpy of fusion

= (5.2 /18) x6.01x 1000J

= 1736.22 J

q2 = mass x specific heat x difference in temperature

= 5.2 g x 4.18J/g.K x (T -273)K

= [21.736 T- 5933.93 ] J

q3 =- mass x specific heat of water x difference in temperature

= - 250 x 4.18 [ T-303]

= {- 1045T + 316635 ] J

equating

1736.22 J + [21.736 T- 5933.93 ] J = {- 1045T + 316635 ] J

Solving for T we get

T= 300.76 K

Now the heat tanferred in the process = {- 1045x 300.76 + 316635 ] J

= 2340.8 J

As delta S(change in entropy) = q /T

= 2340.8/300.76

=7.78J/K