20 g of ice at 0 oC is added to 50 g of water at 30 oC. The container is insulat

Question

20 g of ice at 0 oC is added to 50 g of water at 30 oC. The container is insulated and has a water equivalent of 20g. Calculate the entropy change for the system and the surroundings. The heat of fusion of water is 6.02 kJ/mol; Cp (water)=4.184 J/K/g, Cp(ice)=2.094 J/K/g.

Explanation / Answer

Calculate the final temperature when a 18.8 gram sample of ice at 0oC is placed into a styrofoam cup containing 119 grams of water at 70.3 oC. Assume that there is no loss or gain of heat from the surroundings. Heat of fusion of ice = 333 Jg-1 Specific heat of water = 4.184 JK-1g-1 Final temperature = 49.9 oC Calculate the entropy change that occurs during this process. Entropy change = 5.54 JK-1 Mi = 18.8 gr Ti = 0 deg. C = 273.15 K Mw = 119 gr Tw = 70.3 deg. C = 343.45 K Hfus = 333 J/g.K C = 4.184 J/g.K Energy balance for ice, ?Ei = Q-W, W = 0 ?Ei = Q --> absorb heat Energy balance for ice, ?Ew = -Q --> release heat Combine both, ?Ei + ?Ew = 0, ?E = Ep+Ek+?U, Ep = Ek = 0 ?Ui + ?Uw = 0 ?Ui = -?Uw Mi*(Hfus + C*Tf) = Mw*C*(Tw-Tf) 18.8*(333+4.184Tf) = 119*4.184*(70.3 - Tf) 333+4.184Tf = 26.48383*(70.3-Tf) 333+4.184Tf = 1861.8132 - 26.48383Tf Tf = 49.85 ~ 49.9 deg. C = 323.05 K Entropy change of ice, Tds = dh - vdp, dp~0 ds = dh/T ?Si = Mi*(Hfus/Tfus+ C ln(Tf/Tfus) ?Si = 18.8*(333/273.15 + 4.184*ln(323.05/273.15)) ?Si = 36.117189 J/K Entropy change of water, ?Sw = Mw*C ln(Tf/Tw) ?Sw = 119*4.184*ln(323.05/343.45) ?Sw = -30.4884 J/K Entropy change of system, ?S = ?Si+?Sw ?S = 36.117189 - 30.4884 ?S = 5.63 J/K

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