Entropy as a State Function · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Entropy changes depend on the states

Science · Thermal & Statistical Physics · ages 19-21
Name ______________________   Date ____________

  1. A working substance returns to its starting entropy after a reversible cycle.

    Circle one:   True   False

  2. A system exchanges heat Q reversibly at constant temperature T. Which expression gives its entropy change?

    • Q × T, with T in kelvin
    • Q / T, with T in kelvin
    • T / Q, with T in kelvin

  3. A system releases heat reversibly at constant positive temperature. What is the sign of its entropy change?

    • Positive
    • Zero always
    • Negative

  4. In a model, ice melts at 273 K and takes 2730 J from a reservoir held at 300 K. The water stays at 273 K. What is the combined entropy change of the ice and reservoir?

    • It rises by 0.90 J/K.
    • It falls by 0.90 J/K.
    • It stays at 0 J/K.

  5. A gas held at a constant 300 K absorbs 30 J reversibly. Give its entropy change in J/K.

    Answer: ______________

  6. Which reversible-path result establishes that entropy change is the same between two given equilibrium states?

    • The reversible heat sum is zero along every path between those states.
    • The reversible heat/T sum is zero around any closed loop.
    • The temperature is fixed along every path between the two states.

  7. A solid with constant heat capacity C = 10 J/K warms from 300 K to 600 K without a phase change. Give its entropy change in J/K to two decimal places. Use ln 2 = 0.693.

    Answer: ______________

  8. A system's entropy fell. This fact alone proves that the second law was violated.

    Circle one:   True   False

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Entropy changes depend on the states W1-mt_bg_K5nU-GZ-s1

  1. True · The cycle restores its initial state. Its final entropy equals its initial entropy, so the net change is zero.
  2. Q / T, with T in kelvin · For reversible heat exchange at constant temperature, delta S = Q/T, measured in J/K.
  3. Negative · Released heat has Q < 0. Dividing by a positive temperature in kelvin gives a negative entropy change.
  4. It rises by 0.90 J/K. · The ice gains 2730/273 = 10 J/K. The reservoir loses 2730/300 = 9.1 J/K. Their total change is +0.90 J/K.
  5. 0.1 · Absorption gives Q = +30 J. Thus delta S = 30/300 = +0.1 J/K.
  6. The reversible heat/T sum is zero around any closed loop. · One route forward and another backward form a closed reversible loop. A zero loop sum makes their forward heat/T sums equal.
  7. 6.93 · Delta S = C ln(Tf/Ti) = 10 ln(600/300). Using the supplied ln 2 = 0.693 gives 6.93 J/K.
  8. False · Its surroundings may gain at least as much entropy as the system loses. The second-law test uses the combined isolated whole.
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