Question

If the change in entropy of the surroundings for a process at 451 k and constant pressure is -326 j/k, what is the heat flow absorbed (in kj) by the system?

Answer 1

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:

brainly.com/question/14131507

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