In: Chemistry
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)
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