What is ΔSsurr when 40kJ of heat is removed from the system at 200K?Group of answer choices200 J/K- 0.2 J/K- 200 J/K0.2 J/K
Question
What is ΔSsurr when 40kJ of heat is removed from the system at 200K?
Group of answer choices
- 200 J/K
- -0.2 J/K
- 200 J/K
- 0.2 J/K
Solution
The change in entropy (ΔS) of the surroundings (surr) can be calculated using the formula:
ΔSsurr = -qsys/T
where:
- qsys is the heat absorbed or released by the system (in J), and
- T is the temperature (in K).
In this case, 40 kJ of heat is removed from the system, so qsys = -40,000 J (since 1 kJ = 1,000 J). The temperature is 200 K.
Substituting these values into the formula gives:
ΔSsurr = -(-40,000 J) / 200 K = 200 J/K
So, the change in entropy of the surroundings when 40 kJ of heat is removed from the system at 200 K is 200 J/K.
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