A cross country skier decides to use a reusable heat pack that utilizes polymer

Question

A cross country skier decides to use a reusable heat pack that utilizes polymer crystallization to warm their hand on a cold day outside (the air temperature is -3.9 degree C and remains constant). The pack contains 20 moles of in solution (the heat capacity of this polymer once crystallized is 137.0 J/mol degree C) The skier lost their glove on a fun, but wicked hill and grabbed the heat pack out of a pocket and activated their The pack very quickly heats up to 350 degree C and the skier places it in contact with their cold hand (which has dropped to 20.0 degree C). Given that only 60% the heat from the pack is used the hand, and once the pack is activated they polymer stays crystalline: answer the following questions: What is the final temperature reached by the hand and pack? (Assume the hand is 200g total with a specific heat of 3.47 J/g degree C). What is the entropy change in the hand? What is the entropy change in the pack? What is the entropy change in the surrounding air? What is the total entropy change? Is this process spontaneous?

Explanation / Answer

a)

Tfinal for handpack

Cpolyner = 137 J/mol

mol polymer = 20

Tpack = 35°C

Tair = -3.9°C

Thand = 20°C

60% is transferred

Qhand = mhand*Chand*(Tf-thand)

Qpolym = npolym*Cpolym*(Tf-Tpol)

Qhand = -Qpolym*0.60

mhand*Chand*(Tf-thand) = - npolym*Cpolym*(Tf-Tpol)*0.60

200*3.47*(Tf-20) = -20*137*(Tf-35)*0.60

200*3.47/(-20*137*0.60) * (Tf-20) = (Tf-35)

-0.422*Tf + 20*0.422 = Tf - 35

-1.422*Tf = -35 - 20*0.422

Tf = ( -35 - 20*0.422) /(-1.422) = 30.45°C

b)

dSpolyem = Qpolym/dT = npolym*Cpolym*ln(Tf/Tpol) = 20*137*ln((30+273)/(273+35)) = --44.84J/K

dShand = Qhand/dT = mhand*Chand*ln(Tf/Tpol) = 200*3.47*ln((30+273)/(20+273)) )= 23.290J/K

C)

dSair = Qair / T

Qair = 40% of lost heat

Qair = - npolym*Cpolym*(Tf-Tpol)*0.40

Qair = - 20*137*(30-35)*0.4 = 5480 J

so

dSair = Qair / T = (5480)/(-3.9+273)

dS air = 20.364 J/K

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