In an isothermal and reversible process,945 J of heat is removed from a system and transferred to the surroundings.The temperature is 314 K.What is the change in entropy of the system?

Question

Quizplus · Accepted Answer

For an isothermal and reversible process, the change in entropy ($\Delta S$) can be calculated using the formula $\Delta S = \frac{q_{rev}}{T}$, where $q_{rev}$ is the heat involved in the process and $T$ is the temperature in Kelvin. Given that 945 J of heat is removed from the system, $q_{rev} = -945$ J, and the temperature $T = 314$ K. Thus, $\Delta S = \frac{-945}{314} = -3.01$ J/K.

In an Isothermal and Reversible Process,945 J of Heat Is