CLASSICAL MECHANICS & THERMODYNAMICS - Chapter 8, Exercise 2 Solution
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Calculating Entropy Change (dS = Q/T)

PROBLEM
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400 J of heat is transferred to a system at a constant temperature of
320 K, in a reversible process. Using dS = Q/T, calculate the change
in entropy.

SOLUTION
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Using the entropy change formula for a reversible process at constant
temperature:

    dS = Q / T

Substitute the given values:

    Q = 400 J
    T = 320 K

    dS = 400 / 320
    dS = 1.25 J/K

ANSWER: The system's entropy increases by 1.25 J/K.

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WHY THIS WORKS AS AN ANSWER

Because the process is reversible and happens at a single, constant
temperature, the formula dS = Q/T applies directly without needing
calculus to handle a changing T. The result, 1.25 J/K, has units of
joules per kelvin - entropy is fundamentally different from energy
(measured in plain joules); it measures how that energy transfer is
"spread" relative to the temperature at which it occurs. Transferring
the same 400 J of heat at a LOWER temperature would produce a LARGER
entropy change (since dividing by a smaller T gives a bigger result) -
a real, quantitative reflection of the fact that the same amount of
heat has a bigger disordering effect at a lower temperature than at a
higher one.