Question

Question 18 The entropy change of the surroundings, $\Delta S_{surr}$, is related to heat transfer, $q$, with respect to the system and temperature $T$ by $\circ$ -q/T$_{surr}$ = $\Delta S_{surr}$. $\circ$ +q/T$_{sys}$ = $\Delta S_{surr}$. $\circ$ None of these, unless the system undergoes a reversible process. $\circ$ -q/T$_{sys}$ = $\Delta S_{surr}$ $\circ$ q/T$_{surr}$ = $\Delta S_{surr}$.

          Question 18
The entropy change of the surroundings, $\Delta S_{surr}$, is related to heat transfer, $q$, with respect to the system and temperature $T$ by
$\circ$ -q/T$_{surr}$ = $\Delta S_{surr}$.
$\circ$ +q/T$_{sys}$ = $\Delta S_{surr}$.
$\circ$ None of these, unless the system undergoes a reversible process.
$\circ$ -q/T$_{sys}$ = $\Delta S_{surr}$
$\circ$ q/T$_{surr}$ = $\Delta S_{surr}$.

Question 18
The entropy change of the surroundings, Δ Ssurr, is related to heat transfer, q, with respect to the system and temperature T by
∘ -q/Tsurr = Δ Ssurr.
∘ +q/Tsys = Δ Ssurr.
∘ None of these, unless the system undergoes a reversible process.
∘ -q/Tsys = Δ Ssurr
∘ q/Tsurr = Δ Ssurr.

Added by Kathy R.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

06/29/2023

Step 1

Step 1: The entropy change of the surroundings is equal to the heat transfer divided by the temperature of the surroundings. Show more…

Show all steps

Thanks for your feedback!

Question 18 The entropy change of the surroundings, $\Delta S_{surr}$, is related to heat transfer, q, with respect to the system and temperature T by -q/Tsurr = $\Delta S_{surr}$. +q/Tsys = $\Delta S_{surr}$. None of these, unless the system undergoes a reversible process. -q/Tsys = $\Delta S_{surr}$ q/Tsurr = $\Delta S_{surr}$.