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

Question

A spring is placed in a large thermostat at 27 degree C, and stretched isothermally and reversibly from its equilibrium length L0 to 10 L0. During the reversible stretching 1.00 J of heat are absorbed by the spring. The stretched spring, still in the thermostat, is then released without any restraining tension and allowed to jump back to its original length L0. During this last process, the spring evolves 2.50 J of heat. a) What is the entropy change for the stretching of the spring? b) What is the entropy change for the collapse of the spring? c) What is the entropy change of the universe for the total process

Explanation / Answer

(a) Entropy change during stretching of spring = dq(L-Lo)/T

with constant temperature

Entropy = 1.00 J x 10 = 10 J

(b) Entropy change for collapse of spring at constant temperature = -2.5 J x -10 = 250 J

(c) Entropy change of the universe = -250 + 10 = -240 J

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