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A solid metallic cube of heat capacity S is at temperature 300 K. It is brought in contact with a reservoir at 600 K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the reservoir after reaching the thermal equilibrium is ____
  • a)
    0.69 S
  • b)
    0.54 S
  • c)
    0.27 S
  • d)
    0.19 S
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A solid metallic cube of heat capacity S is at temperature 300 K. It i...
Heat taken by cube=S (600-300)=300S
entropy change of reservoir =-300S÷600=-0.5S
entropy change=S ln (600÷300)=0.69S
hence change in entropy universe= 0.69S-0.5S=0.19S
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Community Answer
A solid metallic cube of heat capacity S is at temperature 300 K. It i...
Entropy Change of the Reservoir

Given:
Temperature of the cube, T1 = 300 K
Temperature of the reservoir, T2 = 600 K

To find:
Entropy change of the reservoir, ΔSreservoir = ?

Entropy Change Calculation:
The entropy change of a reservoir can be calculated using the equation:

ΔSreservoir = Q/T

where ΔSreservoir is the entropy change of the reservoir, Q is the heat transfer between the reservoir and the cube, and T is the temperature of the reservoir.

Heat Transfer Calculation:
The heat transfer between the reservoir and the cube can be calculated using the equation:

Q = m * c * ΔT

where Q is the heat transfer, m is the mass of the cube, c is the specific heat capacity of the cube, and ΔT is the change in temperature.

Since the cube is solid and metallic, its specific heat capacity is given as S.

ΔT = T2 - T1 = 600 K - 300 K = 300 K

Substituting the values into the equation, we get:

Q = m * S * ΔT

Entropy Change of the Reservoir Calculation:
Now, substituting the value of Q into the entropy change equation, we get:

ΔSreservoir = Q / T2
= (m * S * ΔT) / T2

Since the cube and the reservoir reach thermal equilibrium, their temperatures become equal. Therefore, the final temperature of the cube is also 600 K.

ΔSreservoir = (m * S * ΔT) / T2
= (m * S * ΔT) / T2
= (m * S * 300 K) / 600 K
= (m * S * 0.5)

Since the heat capacity of the cube is given as S, the entropy change of the reservoir is equal to 0.5 times the heat capacity of the cube.

ΔSreservoir = 0.5 * S

Hence, the correct answer is option D) 0.19 S.
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A solid metallic cube of heat capacity S is at temperature 300 K. It is brought in contact with a reservoir at 600 K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the reservoir after reaching the thermal equilibrium is ____a)0.69 Sb)0.54 Sc)0.27 Sd)0.19 SCorrect answer is option 'D'. Can you explain this answer?
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A solid metallic cube of heat capacity S is at temperature 300 K. It is brought in contact with a reservoir at 600 K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the reservoir after reaching the thermal equilibrium is ____a)0.69 Sb)0.54 Sc)0.27 Sd)0.19 SCorrect answer is option 'D'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A solid metallic cube of heat capacity S is at temperature 300 K. It is brought in contact with a reservoir at 600 K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the reservoir after reaching the thermal equilibrium is ____a)0.69 Sb)0.54 Sc)0.27 Sd)0.19 SCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid metallic cube of heat capacity S is at temperature 300 K. It is brought in contact with a reservoir at 600 K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the reservoir after reaching the thermal equilibrium is ____a)0.69 Sb)0.54 Sc)0.27 Sd)0.19 SCorrect answer is option 'D'. Can you explain this answer?.
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