Question

A eco-conscious dad notices that we left the computer and light on from the last homework. After scolding us about the electrical bill, he turns them both off, and they eventually cool to the ambient temperature of 25 ^oC.

a) What is the total change in entropy for the computer chip (or bulb) for this cooling process? Find the total entropy change for the universe. (Recall that the chip is 50. g silicon, 20. g copper, and 65 g. polyvinyl chloride, initially running at a temperature of 75 ^oC.)

b) Find the specific entropy change for the light bulb in this cooling process. Calculate the specific entropy generated for the bulb. (The light bulb is initially filled with argon, an ideal gas, at a temperature of 65.0 ^oC a volume of 40. cm³ and a pressure of 103.8 kPa. Neglect the glass and metal comprising the rest of the light bulb.)

Answer 1

Tsys = 273+75 = 348 K

Tsurr = 273 + 25 = 298 K

dT = 25 - 75 = - 50 DegC

Total heat evolved by the computer chip to the surrounding,

Q = mxS(silicon)xdT + mxS(Copper)xdT + mxS(pvc)xdT

= 50.0gx(0.710 Jg^-1DegC^-1)x(- 50 DegC) + 20.0x(0.384 Jg^-1​DegC^-1)x(- 50 DegC)

+ 65.0gx(0.90 Jg^-1DegC^-1)x(- 50 DegC)

= - 5084 J

Ta = (348+298) / 2 = 323K

Now entropy change for the computer chip (DeltaS) can be calculated from the following formulae

DeltaS(sys) = Q(sys) / Ta = (- 5084 J) / (323) K = - 15.74 J/K (answer)

Q(surr) = - (- 5084 J ) = + 5084 J

T(surr) = 273+25 = 298 K

DeltaS(surr) = Q(surr) / Tsurr = (+ 5084 J) / 298K = + 17.06 J/K

DeltaS(universe) = DeltaS(sys) + DeltaS(surr) = - 15.74 J/K + 17.06 J/K = 1.32  J/K (answer)

(b) Moles of Ar in bulb, n = PV/RT = 103.8x10³Pa x 40.0 x 10^-6 m³ / 8.314 Jmol^-1K^-1 x (273+65)

= 0.00148 mol Ar

Mass of Ar = 0.00148 mol x 39.95 g/mol = 0.0591 g

Q = m x S(Ar)xdT = 0.0591 g x 0.522J/g.DegC x (65 - 25) = 1.234 J

Ta = (298 +338)/2 K = 318 K

DeltaS = Q /Ta = 1.234 J / 318 K = 0.00388 J/K (answer)