به انگلیسی

Entropy changes depend on the states

Science · Thermal & Statistical Physics · ages 19-21

Name ______________________ Date ____________

A working substance returns to its starting entropy after a reversible cycle.
Circle one: True False
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
A system releases heat reversibly at constant positive temperature. What is the sign of its entropy change?
- Positive
- Zero always
- Negative
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.
A gas held at a constant 300 K absorbs 30 J reversibly. Give its entropy change in J/K.
Answer: ______________
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.
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: ______________
A system's entropy fell. This fact alone proves that the second law was violated.
Circle one: True False

Answer key

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

True · The cycle restores its initial state. Its final entropy equals its initial entropy, so the net change is zero.
Q / T, with T in kelvin · For reversible heat exchange at constant temperature, delta S = Q/T, measured in J/K.
Negative · Released heat has Q < 0. Dividing by a positive temperature in kelvin gives a negative entropy change.
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.
0.1 · Absorption gives Q = +30 J. Thus delta S = 30/300 = +0.1 J/K.
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.
6.93 · Delta S = C ln(Tf/Ti) = 10 ln(600/300). Using the supplied ln 2 = 0.693 gives 6.93 J/K.
False · Its surroundings may gain at least as much entropy as the system loses. The second-law test uses the combined isolated whole.