A spring is placed in a large thermostat at 27 degree C and stretched isothermal
ID: 3279172 • Letter: A
Question
A spring is placed in a large thermostat at 27 degree C and stretched isothermally and reversibly from its equilibrium length, L_0, to 5L_0. During this process 1.0 J of heat is absorbed by the spring. The stretched spring, still in the thermostat, is then released without any restraining back tension and snaps irreversibly back to its equilibrium length L_0. During the contraction process 2.5 J of heat is lost from the spring to the thermostat. Calculate: a. The overall delta E (stretching plus contraction) for the spring b. The work done on the spring in the stretching process C. The value of delta S for the spring in the stretching step d. The value of delta S for the spring in the contraction step e. The value of delta S for the spring plus the surroundings for the overall process f. Does this process violate any laws of thermodynamics? Justify your answer.Explanation / Answer
given, temperature of the apparatus = T = 27 C
strech = L to 5L
heat absorbed in streching = Ho = 1 J
heat relesed during snapping back, H = 2.5 J
a. dE for the spring = heat absored during expansion - heat released during the contraction = 1 - 2.5 = -1.5 J
b. let the work done on the spring be W
then W + Ho = H
W = H -Ho = 2.5 - 1 = 1.5 J
c. dS = dQ/T
dQ = 1 J for streching
T = 27 C = 27 + 273.16 = 300.16 K
dS = 1/300.16 = 3.33155*10^-3 J/K
d. for the contraction step
dQ = 2.5 J
T = 27 C = 300.16 K
dS = 2.5/300.16 = 8.32889*10^-3 j/k
E. for the overall process
dQ for the spring = -1 + 2.5 = 1.5 J given to the surroundings
so dQ for spring plus surroundings = 0
hence dS = 0
F. this proces does not violate laws of thermo dynamics because
heat input = change in internalenergy + work done by the equipment
hence the laws of thermodynamics are foll;owed by the system
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