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What is the change in entropy of the system when 1.25 " 10^{3}J of heat is added to the system that is maintained at a constant 75.0 C?

My work was

[itex]\begin{align*}\Delta S &= \frac{Q}{T} \\ \Delta S &= \frac{1.25 \times 10^3 \, \mathrm{J}}{348 \, \mathrm{K}} \\ \Delta S &= 3.59 \,\,\mathrm{\left( J/K\right )} \end{align*}[/itex]

but I was stuck in between two answer choices

c) + 3.59 J/K

d) # 3.59 J/K

I had no idea what the hashtag meant, so I skipped it. Can someone clear this up for me?