A spring is placed in a large thermostat at 27 degree C and stretched isothermal

Question

A spring is placed in a large thermostat at 27 degree C and stretched isothermally and from its equilibrium length L During this reversible stretching. 1 00 J absorbed b the spring The stretched spring, still in the constant-temperature released without ans restraining back-tension and is allowed to jump back to its initial l (T 'S During this spontaneous process, the spring evolves 2 50 J of heat What is the entropy change for the stretching of the spring ? What is the entropy change for the collapse of the spring'1 Calculate What is the entropy change for the universe (spring plus process, stretching plus return collapse to initial L surrounding thermostat J for the total FI (Entropy with phase transitions) Water droplets in the atmosphere persist as super liquids (in other words, they do not freeze) until -35 degree C Consider one mole of water' frat r placed in surroundings at -35 degree C At first it does not freeze (it remains as supercooled water equilibrated at the temperature of the surroundings! Suddenly it freezes " a. Calculate the entropy change in the system during the freezing, making use of the following data: Calculate the entropy change in the surroundings. Assume the surroundings are sufficiently large that the temperature of the surroundings is constant. Calculate the total entropy change in the system and surroundings Is this process spontaneous? Why or why not.

Explanation / Answer

S = dQrev/T

To make the process reversible we can take the following path:

1) Take the supercooled water up to 0°C reversibly.

2) Freeze the water (at 1 atm, 0°C)

3) Take the ice back down to -35°C reversibly.

T1 = -35 + 273 = 238 K

T2 = 0 + 273 = 273 K

a) n = 1

Ssystem = nCps* log(T2/T1) - nHfus/T2 + nCpw* log(T1/T2)

= 1* (75.3 log(273/238) - 6.02*103/273 + 37.7 log(238/273))

= 10.33 - 22.05 - 5.17 = -16.89 J/K

b) Tsurr = -35 + 273 = 238 K

Ssurr = - (nCps* (T2-T1)/Tsurr - nHfus/Tsurr+ nCpw* (T1-T2)/Tsurr )

= -1* (37.7*35 - 6.02*103 + 75.3* -35) / 238

= 19.76 J/K

c) Stotal =Ssystem + Ssurr

= -16.89 + 19.76 = 2.87 J/K > 0

Process is spontaneous as Stotal is greater than 0.

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