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When 100 g of water is reversibly heated from 50 degree celsius to 75 degree celsius at 1 atm the change in entropy of the universe is?
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When 100 g of water is reversibly heated from 50 degree celsius to 75 ...
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This question doesn’t specify the water’s final state (100 C vapor or liquid?) and assigns an incorrect value to liquid water’s specific heat ( “heat capacity”) - it’s 4180 J per kg, not mole
The liquid’s entropy change would be its heat capacity (Cp) times the natural log of the two absolute temperatures or 4.18 J/g degree K*18 g/mol *(ln*((273+100)/(273+90)= 2.0447 J/(K*mol)
For 10 kg (10000/18 moles) of water, that ΔS works out to 1136 J/K
Since it takes about 540 calories (or 4.18*540 J) to evaporate 1 gram of water at atmospheric pressure and all of that energy goes into increasing its entropy, that change would add another 60,500 (10,000*(4.18*540/373) J/K
Consequently if the water ends up as steam. its total entropy change would 1136+60500 or 61700 J/degree.
The liquid’s entropy change would be its heat capacity (Cp) times the natural log of the two absolute temperatures or 4.18 J/g degree K*18 g/mol *(ln*((273+100)/(273+90)= 2.0447 J/(K*mol)
For 10 kg (10000/18 moles) of water, that ΔS works out to 1136 J/K
Since it takes about 540 calories (or 4.18*540 J) to evaporate 1 gram of water at atmospheric pressure and all of that energy goes into increasing its entropy, that change would add another 60,500 (10,000*(4.18*540/373) J/K
Consequently if the water ends up as steam. its total entropy change would 1136+60500 or 61700 J/degree.
This question is part of UPSC exam. View all IIT JAM courses
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When 100 g of water is reversibly heated from 50 degree celsius to 75 ...
Calculation of Change in Entropy of the Universe
To calculate the change in entropy of the universe during the reversible heating of water from 50 degrees Celsius to 75 degrees Celsius at 1 atm, we need to consider the entropy change of both the system (water) and the surroundings.
Entropy Change of the System
The entropy change of the system can be calculated using the equation:
ΔS_system = mcΔT
Where:
ΔS_system = change in entropy of the system
m = mass of water (100 g)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (75°C - 50°C = 25°C)
Substituting the values into the equation:
ΔS_system = (100 g)(4.18 J/g°C)(25°C)
ΔS_system = 10,450 J/°C
Entropy Change of the Surroundings
The entropy change of the surroundings can be calculated using the equation:
ΔS_surroundings = -ΔH/T
Where:
ΔS_surroundings = change in entropy of the surroundings
ΔH = change in enthalpy of the system (heat transferred)
T = temperature in Kelvin (75°C + 273 = 348 K)
Since the process is reversible, the heat transferred can be calculated using the equation:
ΔH = mcΔT
Substituting the values into the equation:
ΔH = (100 g)(4.18 J/g°C)(25°C)
ΔH = 10,450 J
Now, substituting the values of ΔH and T into the equation for ΔS_surroundings:
ΔS_surroundings = -(10,450 J)/(348 K)
ΔS_surroundings = -30 J/K
Total Change in Entropy of the Universe
The total change in entropy of the universe is the sum of the entropy change of the system and the surroundings:
ΔS_universe = ΔS_system + ΔS_surroundings
ΔS_universe = 10,450 J/°C - 30 J/K
Since the units of the entropy change must be consistent, we convert J/°C to J/K:
ΔS_universe = 10,450 J/K - 30 J/K
ΔS_universe = 10,420 J/K
Therefore, the change in entropy of the universe during the reversible heating of water from 50 degrees Celsius to 75 degrees Celsius at 1 atm is 10,420 J/K.
To calculate the change in entropy of the universe during the reversible heating of water from 50 degrees Celsius to 75 degrees Celsius at 1 atm, we need to consider the entropy change of both the system (water) and the surroundings.
Entropy Change of the System
The entropy change of the system can be calculated using the equation:
ΔS_system = mcΔT
Where:
ΔS_system = change in entropy of the system
m = mass of water (100 g)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (75°C - 50°C = 25°C)
Substituting the values into the equation:
ΔS_system = (100 g)(4.18 J/g°C)(25°C)
ΔS_system = 10,450 J/°C
Entropy Change of the Surroundings
The entropy change of the surroundings can be calculated using the equation:
ΔS_surroundings = -ΔH/T
Where:
ΔS_surroundings = change in entropy of the surroundings
ΔH = change in enthalpy of the system (heat transferred)
T = temperature in Kelvin (75°C + 273 = 348 K)
Since the process is reversible, the heat transferred can be calculated using the equation:
ΔH = mcΔT
Substituting the values into the equation:
ΔH = (100 g)(4.18 J/g°C)(25°C)
ΔH = 10,450 J
Now, substituting the values of ΔH and T into the equation for ΔS_surroundings:
ΔS_surroundings = -(10,450 J)/(348 K)
ΔS_surroundings = -30 J/K
Total Change in Entropy of the Universe
The total change in entropy of the universe is the sum of the entropy change of the system and the surroundings:
ΔS_universe = ΔS_system + ΔS_surroundings
ΔS_universe = 10,450 J/°C - 30 J/K
Since the units of the entropy change must be consistent, we convert J/°C to J/K:
ΔS_universe = 10,450 J/K - 30 J/K
ΔS_universe = 10,420 J/K
Therefore, the change in entropy of the universe during the reversible heating of water from 50 degrees Celsius to 75 degrees Celsius at 1 atm is 10,420 J/K.
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When 100 g of water is reversibly heated from 50 degree celsius to 75 degree celsius at 1 atm the change in entropy of the universe is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about When 100 g of water is reversibly heated from 50 degree celsius to 75 degree celsius at 1 atm the change in entropy of the universe is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 100 g of water is reversibly heated from 50 degree celsius to 75 degree celsius at 1 atm the change in entropy of the universe is?.
When 100 g of water is reversibly heated from 50 degree celsius to 75 degree celsius at 1 atm the change in entropy of the universe is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about When 100 g of water is reversibly heated from 50 degree celsius to 75 degree celsius at 1 atm the change in entropy of the universe is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 100 g of water is reversibly heated from 50 degree celsius to 75 degree celsius at 1 atm the change in entropy of the universe is?.
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