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1000gm of water at temperature 300K is isobarically and adiabatically mixed with an equal mass of water at 500K. C = 0.01 cal K-1gm-1 . The entropy change of the universe in (cal K-1 ) is
Correct answer is '0.645'. Can you explain this answer?
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1000gm of water at temperature 300K is isobarically and adiabatically ...
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1000gm of water at temperature 300K is isobarically and adiabatically ...
Solution:
Given:
Mass of water, m = 1000 gm
Temperature of water 1, T1 = 300 K
Temperature of water 2, T2 = 500 K
Specific heat capacity of water, C = 0.01 cal K-1gm-1
To find:
The entropy change of the universe, ΔS
We know that the change in entropy of a system can be calculated using the formula:
ΔS = mcΔT / T
where,
m is the mass of the system
c is the specific heat capacity of the system
ΔT is the change in temperature of the system
T is the temperature of the system
1. Calculate the change in temperature:
ΔT = T2 - T1
= 500 K - 300 K
= 200 K
2. Calculate the entropy change of the system:
ΔS = mcΔT / T
Substituting the given values:
ΔS = (1000 gm)(0.01 cal K-1gm-1)(200 K) / 300 K
= 2000 cal K / 300 K
= 6.67 cal K
3. Calculate the entropy change of the surroundings:
Since the process is adiabatic, there is no heat exchange between the system and the surroundings. Therefore, the entropy change of the surroundings is zero.
4. Calculate the entropy change of the universe:
ΔSuniverse = ΔSsystem + ΔSsurroundings
= 6.67 cal K + 0 cal K
= 6.67 cal K
Therefore, the entropy change of the universe is 6.67 cal K.
Given:
Mass of water, m = 1000 gm
Temperature of water 1, T1 = 300 K
Temperature of water 2, T2 = 500 K
Specific heat capacity of water, C = 0.01 cal K-1gm-1
To find:
The entropy change of the universe, ΔS
We know that the change in entropy of a system can be calculated using the formula:
ΔS = mcΔT / T
where,
m is the mass of the system
c is the specific heat capacity of the system
ΔT is the change in temperature of the system
T is the temperature of the system
1. Calculate the change in temperature:
ΔT = T2 - T1
= 500 K - 300 K
= 200 K
2. Calculate the entropy change of the system:
ΔS = mcΔT / T
Substituting the given values:
ΔS = (1000 gm)(0.01 cal K-1gm-1)(200 K) / 300 K
= 2000 cal K / 300 K
= 6.67 cal K
3. Calculate the entropy change of the surroundings:
Since the process is adiabatic, there is no heat exchange between the system and the surroundings. Therefore, the entropy change of the surroundings is zero.
4. Calculate the entropy change of the universe:
ΔSuniverse = ΔSsystem + ΔSsurroundings
= 6.67 cal K + 0 cal K
= 6.67 cal K
Therefore, the entropy change of the universe is 6.67 cal K.
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1000gm of water at temperature 300K is isobarically and adiabatically mixed with an equal mass of water at 500K. C = 0.01 cal K-1gm-1 . The entropy change of the universe in (cal K-1 ) isCorrect answer is '0.645'. Can you explain this answer?
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1000gm of water at temperature 300K is isobarically and adiabatically mixed with an equal mass of water at 500K. C = 0.01 cal K-1gm-1 . The entropy change of the universe in (cal K-1 ) isCorrect answer is '0.645'. 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 1000gm of water at temperature 300K is isobarically and adiabatically mixed with an equal mass of water at 500K. C = 0.01 cal K-1gm-1 . The entropy change of the universe in (cal K-1 ) isCorrect answer is '0.645'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1000gm of water at temperature 300K is isobarically and adiabatically mixed with an equal mass of water at 500K. C = 0.01 cal K-1gm-1 . The entropy change of the universe in (cal K-1 ) isCorrect answer is '0.645'. Can you explain this answer?.
1000gm of water at temperature 300K is isobarically and adiabatically mixed with an equal mass of water at 500K. C = 0.01 cal K-1gm-1 . The entropy change of the universe in (cal K-1 ) isCorrect answer is '0.645'. 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 1000gm of water at temperature 300K is isobarically and adiabatically mixed with an equal mass of water at 500K. C = 0.01 cal K-1gm-1 . The entropy change of the universe in (cal K-1 ) isCorrect answer is '0.645'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1000gm of water at temperature 300K is isobarically and adiabatically mixed with an equal mass of water at 500K. C = 0.01 cal K-1gm-1 . The entropy change of the universe in (cal K-1 ) isCorrect answer is '0.645'. Can you explain this answer?.
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