An insulated Thermos contains 130 g of water at 80.0°C. You put in a 12.0 g ice

Question

An insulated Thermos contains 130 g of water at 80.0°C. You put in a 12.0 g ice cube at 0°C to form a system of ice + original water. (a) What is the equilibrium temperature of the system? What are the entropy changes of the water that was originally the ice cube (b) as it melts and (c) as it warms to the equilibrium temperature? (d) What is the entropy change of the original water as it cools to the equilibrium temperature? (e) What is the net entropy change of the ice + original water system as it reaches the equilibrium temperature?

Explanation / Answer

(a)
Specific Heat of Water, Cw = 4.186 J/gm k
Latent Heat of ice, Li = 333.55 J/gm
Mass of ice , mi = 12.0 g
Mass of Water, mw = 130 g
Let the Final Temperature = Tf

Heat Lost by Water = Heat gained by ice
mw * cw * T = mi * Lf + mi * cw * T
130 * 4.186 * (80 - Tf) = 12 * 333.55 + 12 * 4.186 * Tf
Solving for Tf
Tf = 66.5 °C
Equilibrium Temperature of the system, Tf = 66.5 °C


T1 = 273 K
T2 = 339.5 K
T3 = 353 K

Entropy Changes -

(b)
S_ice = (Mi*Lf)/T_avg
S_ice = (12 * 333.55) / 273
S_ice = 14.66 J/k


(c)
S_ice = (Mi*Cw*T)/T_avg
S_ice = (12*4.186*(66.5 - 0))/((339.5 + 273)/2)
S_ice = 10.91 J/k

(d)
S_water = (Mw*Cw*T)/T_avg
S_water = ( 130 * 4.186 * (66.5 - 80)) / ((339.5 + 353)/2)
S_water = -21.22 J/k

(e)
Net entropy change of the ice + original water system =
S = 14.66 + 10.91 - 21.22
S = 4.35 J/k

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