You have 100 g of ice at -20 C that is heated to a temperature of 0 C (stage 1),

Question

You have 100 g of ice at -20 C that is heated to a temperature of 0 C (stage 1), melted at 0 C (stage 2), heated to 100 C (stage 3), vaporized at 100 C (stage 4) and heated to 150 C (stage 5). The heat transfer rate into the system is a constant = 1 kJ min-1 and all processes occur at constant pressure. At the end of each stage, calculate the (1) time, (2) temperature, (3) enthalpy, and (4) entropy of the system. You can use an enthalpy and entropy of zero at T = -20 C. Plot the temperature, enthalpy and entropy with time on the same graph or separate graphs. You can assume constant molar heat capacities over the range of temperatures. Use a molar heat capacity of solid phase of water of (Cpm)ice = 36 J mol-1 K-1

Explanation / Answer

From -20 deg.c to 0 deg.c, it is sensible heat and is given by= moles* specific heat* temperature difference =(100/18)*36*(0+20) Joules=4000 Joules= 4 Kj

Heating rate= 1 Kj/min, time = 4/1= 4 min

Entropy change= Moles*specific heat*ln(T2/T1)= (100/18)*36* ln {(273/(-20+273)}=15.21

Since entropy at -20 deg.c is zero, entropy = 15.21 J/K

Temperature at the end of the process= 0 deg.c

2. at 0 deg.c, ice is supplied with latent heat of fusion whose value is =334 J/gm

Hence enthalpy change = mass* specific enthalpy= 100*334 =33400 joules= 33.4 Kj

Time required= 33.4/1= 33.4 min

Entropy change of fusion = 33.4*1000/273

Total Entropy change up to fusion = 15.32+33.4*1000/273 =138 J/K

3. from 0 to 100 deg.c, it is sensible heat = 100*4.184*(100-0)=41840 joules= 41.840 Kj

Time required= 41.84/1= 41.84 min

Entropy change= 100*4.184*ln(373/273)= 131 J/K

Up to this point entropy change= 138+131= 269 J/K

4. at 100 deg.c water receives latent heat which is 2230 j/gm

Enthalpy = 2230*100 joules= 223000 Joules= 223 KJ, tine =223/1= 223 min

Entropy = 223000/373 J/K=598 J/K

Total entropy change= 269+598= 867 J/K

5. from 100 deg.c to 150 deg.c it is super heat which is = 100*2*50=10000 Joules ( 2J/gm.deg.c is specific heat of water vapor)= 10 Kj

Time =10/1= 10 min, entropy = 100*2*ln{(150+273)/(100+273)}=25 J/K

Total entropy change =867+25 =892 J/K

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