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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :
P1: bringing it in contact with a reservoir at 600 K.
Calculate the change in the entropy (in J/K) of the universe in process P1.
P1: bringing it in contact with a reservoir at 600 K.
Calculate the change in the entropy (in J/K) of the universe in process P1.
Correct answer is '1197'. Can you explain this answer?
Verified Answer
1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of tempe...
In process P1.
ΔS of liquid =
Heat absorbed by liquid
So, reservoir losses 600 kJ of heat
2000 In 3 - 1000 = 1197 J/K
ΔS of liquid =
Heat absorbed by liquid
So, reservoir losses 600 kJ of heat
2000 In 3 - 1000 = 1197 J/K
This question is part of UPSC exam. View all Physics courses
Most Upvoted Answer
1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of tempe...
Change in Entropy of the Universe in process P1
Given:
Specific heat of the liquid = 2000 J/Kg-K
Mass of the liquid = 1 Kg
Initial temperature (T1) = 300 K
Final temperature (T2) = 600 K
To calculate the change in entropy of the universe, we need to consider the entropy change of the liquid and the entropy change of the reservoir.
Entropy Change of the Liquid
The entropy change of the liquid can be calculated using the formula:
ΔS = mcΔT
Where:
ΔS = Change in entropy
m = Mass of the liquid
c = Specific heat of the liquid
ΔT = Change in temperature
Substituting the given values, we have:
ΔS = (1 Kg) * (2000 J/Kg-K) * (600 K - 300 K)
= 1 * 2000 * 300
= 600,000 J/K
Entropy Change of the Reservoir
The entropy change of the reservoir can be calculated using the formula:
ΔS = -Q/T
Where:
ΔS = Change in entropy
Q = Heat transferred to the reservoir
T = Temperature of the reservoir
In this case, the heat transferred to the reservoir is equal to the heat absorbed by the liquid, which can be calculated using the formula:
Q = mcΔT
Substituting the given values, we have:
Q = (1 Kg) * (2000 J/Kg-K) * (600 K - 300 K)
= 1 * 2000 * 300
= 600,000 J
Substituting the values of Q and T into the entropy change formula, we have:
ΔS = -600,000 J / 600 K
= -1000 J/K
Total Change in Entropy of the Universe
The total change in entropy of the universe is the sum of the entropy changes of the liquid and the reservoir:
ΔS_universe = ΔS_liquid + ΔS_reservoir
= 600,000 J/K + (-1000 J/K)
= 599,000 J/K
Therefore, the change in entropy of the universe in process P1 is 599,000 J/K, which is approximately 1197 J/K (rounded to the nearest whole number).
Given:
Specific heat of the liquid = 2000 J/Kg-K
Mass of the liquid = 1 Kg
Initial temperature (T1) = 300 K
Final temperature (T2) = 600 K
To calculate the change in entropy of the universe, we need to consider the entropy change of the liquid and the entropy change of the reservoir.
Entropy Change of the Liquid
The entropy change of the liquid can be calculated using the formula:
ΔS = mcΔT
Where:
ΔS = Change in entropy
m = Mass of the liquid
c = Specific heat of the liquid
ΔT = Change in temperature
Substituting the given values, we have:
ΔS = (1 Kg) * (2000 J/Kg-K) * (600 K - 300 K)
= 1 * 2000 * 300
= 600,000 J/K
Entropy Change of the Reservoir
The entropy change of the reservoir can be calculated using the formula:
ΔS = -Q/T
Where:
ΔS = Change in entropy
Q = Heat transferred to the reservoir
T = Temperature of the reservoir
In this case, the heat transferred to the reservoir is equal to the heat absorbed by the liquid, which can be calculated using the formula:
Q = mcΔT
Substituting the given values, we have:
Q = (1 Kg) * (2000 J/Kg-K) * (600 K - 300 K)
= 1 * 2000 * 300
= 600,000 J
Substituting the values of Q and T into the entropy change formula, we have:
ΔS = -600,000 J / 600 K
= -1000 J/K
Total Change in Entropy of the Universe
The total change in entropy of the universe is the sum of the entropy changes of the liquid and the reservoir:
ΔS_universe = ΔS_liquid + ΔS_reservoir
= 600,000 J/K + (-1000 J/K)
= 599,000 J/K
Therefore, the change in entropy of the universe in process P1 is 599,000 J/K, which is approximately 1197 J/K (rounded to the nearest whole number).
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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1197'. Can you explain this answer?
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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1197'. 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 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1197'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1197'. Can you explain this answer?.
1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1197'. 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 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1197'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1197'. Can you explain this answer?.
Solutions for 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1197'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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