An 13 g ice cube at -15.0?C is put into a Thermos flask containing 100 cm3 of wa

Question

An 13 g ice cube at -15.0?C is put into a Thermos flask containing 100 cm3 of water at 21.0?C. By how much has the entropy of the cube-water system changed when a final equilibrium state is reached? The specific heat of ice is 2200 J/kg K and that of liquid water is 4187 J/kg K. The heat of fusion of water is 333

Explanation / Answer

Mi = 11 gr Ti = -15 deg. C = 258 K Mw = 120 gr Tw = 21 deg. C = 294 K Hfus = 333 J/g.K Cw = 4.187 J/g.K Ci = 2.22 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*[Ci*(Tfus-Ti) + Hfus + Cw*(Tf-Tfus)] = Mw*C*(Tw-Tf) Solve for final temperature, Tf. Entropy change of ice, Tds = dh - vdp, dp~0 ds = dh/T ?Si = Mi[(Ci*ln(Tfus/Ti) + Hfus/Tfus+ Cw ln(Tf/Tfus] Entropy change of water, ?Sw = Mw*C ln(Tf/Tw) Entropy change of system, ?S = ?Si+?Sw

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