Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
If the change in entropy of the surroundings for a process at 451 k and constant pressure is -326 j/k, then heat flow absorbed (in kj) by the system is -147.026kJ.
What is entropy?
The entropy of particle is defined as how random it move. It shows the randomness of the system or may be disorders of the system. It is used to measure the unavailable energy for performing useful work.
Unit of entropy = J/K
Formula:
∆s = ∆Q/T
where,
∆s = change in entropy of the surrounding = -326J/K
∆Q = heat absorbed from surrounding
T = Temperature = 451K
∆Q = ∆s × T
∆Q = -326 × 451
∆Q = 147,026 J
∆Q = 147.026 kJ
Thus we find that the heat absorbed by the system is 147.026 kJ.
learn more about entropy:
https://brainly.com/question/14131507
#SPJ1
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.