Chemistry I / Entropy and the Second Law
Practice question · Numerical answer

A process releases 4500 J of heat reversibly to surroundings held at 250 K. Compute the entropy change of the SURROUNDINGS in J/K.

Hints
  1. The surroundings absorb the heat the system releases, so their q is positive.
  2. Compute 4500 / 250.
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18 (answers within ±0.2 count)

Why

ΔSsurr=qrev/T=4500/250=+18\Delta S_{surr} = q_{rev}/T = 4500/250 = +18 J/K. The surroundings gain entropy even though the system lost heat, and it is this gain that can license a spontaneous process in which the system’s own entropy falls.

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