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In a process occurring in a closed system F, the heat transferred from F to the surroundings E is 600 J. If the temperature of E is 300 K and that of F is in the range 380 - 400 K, the entropy changes of the surroundings (SSE) and system (ΔSF), in J/K, are given by
  • a)
    ΔSE = 2, ΔSF = -2
  • b)
    ΔSE = -2, ΔSF = 2
  • c)
    ΔSE = 2, ΔSF = -2
  • d)
    ΔSE = 2, ΔSF > -2
Correct answer is option 'D'. Can you explain this answer?
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In a process occurring in a closed system F, the heat transferred from...
SSF) can be calculated using the formula:

SSE = -Q/T
SSF = Q/T

Given:
Q = 600 J
TE = 300 K
TF = 380 - 400 K

To calculate SSE, we use the formula SSE = -Q/TE:

SSE = -600 J / 300 K
SSE = -2 J/K

To calculate SSF, we use the formula SSF = Q/TF:

SSF = 600 J / (380 - 400 K)
SSF = 600 J / -20 K
SSF = -30 J/K

Therefore, the entropy change of the surroundings (SSE) is -2 J/K and the entropy change of the system (SSF) is -30 J/K.
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In a process occurring in a closed system F, the heat transferred from F to the surroundings E is 600 J. If the temperature of E is 300 K and that of F is in the range 380 - 400 K, the entropy changes of the surroundings (SSE) and system (ΔSF), in J/K, are given bya)ΔSE = 2, ΔSF = -2b)ΔSE = -2, ΔSF = 2c)ΔSE = 2, ΔSF = -2d)ΔSE = 2, ΔSF>-2Correct answer is option 'D'. Can you explain this answer?
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