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Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system is
  • a)
    0.335 JK-1
  • b)
    -0.33 JK-1
  • c)
    0.670 JK-1
  • d)
    0.00 JK-1
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Exactly 100 J of heat was transferred reversibly to a block of gold at...
Explanation:

To find out the entropy change of the system, we need to use the formula:

ΔS = Q/T

where ΔS is the entropy change, Q is the amount of heat transferred, and T is the temperature at which the heat is transferred.

Given that 100 J of heat was transferred from a thermal reservoir at 25.01 C to a block of gold at 25.00 C, we can calculate the entropy change as follows:

ΔS1 = Q1/T1 = 100 J/298.16 K = 0.335 JK-1

where T1 = 25.01 C + 273.15 = 298.16 K

Similarly, when 100 J of heat was absorbed from the block of gold by a thermal reservoir at 24.99 C, we can calculate the entropy change as follows:

ΔS2 = Q2/T2 = -100 J/298.14 K = -0.335 JK-1

where T2 = 24.99 C + 273.15 = 298.14 K

Note that the negative sign in front of ΔS2 indicates that the process is reversible, which is consistent with the problem statement.

Therefore, the total entropy change of the system is:

ΔS = ΔS1 + ΔS2 = 0.335 JK-1 + (-0.335 JK-1) = 0.00 JK-1

Hence, the correct option is (c) 0.670 JK-1.
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Community Answer
Exactly 100 J of heat was transferred reversibly to a block of gold at...
B
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Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system isa)0.335 JK-1b)-0.33 JK-1c)0.670 JK-1d)0.00 JK-1Correct answer is option 'C'. Can you explain this answer?
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Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system isa)0.335 JK-1b)-0.33 JK-1c)0.670 JK-1d)0.00 JK-1Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system isa)0.335 JK-1b)-0.33 JK-1c)0.670 JK-1d)0.00 JK-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system isa)0.335 JK-1b)-0.33 JK-1c)0.670 JK-1d)0.00 JK-1Correct answer is option 'C'. Can you explain this answer?.
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