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A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for water
Correct answer is '927'. Can you explain this answer?
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Verified Answer
A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reve...
Mass of water = 1.78kg
Mass of ice = 262g
Mass of ice-water mixture = 1.78 kg + 262g
= 2.04kg
If eventually, ice and and water have the same mass, then the final state will have 1.02kg of each.
Thus, mass of water that changed into ice m will be the difference of mass of water mW and mass of final state mS .
So, m = mW – mS
⇒m = 1.78 – 1.02
= 0.76kg
The change of water at 0ºC to ice at 0ºC is isothermal.
⇒ ΔS = –mL/T
= –(0.76kg) (333×103J/kg) (273K)
= –927 J/K
⇒ Change in entropy of the system during this process will be –927J/K.
The correct answer is: 927
Mass of ice = 262g
Mass of ice-water mixture = 1.78 kg + 262g
= 2.04kg
If eventually, ice and and water have the same mass, then the final state will have 1.02kg of each.
Thus, mass of water that changed into ice m will be the difference of mass of water mW and mass of final state mS .
So, m = mW – mS
⇒m = 1.78 – 1.02
= 0.76kg
The change of water at 0ºC to ice at 0ºC is isothermal.
⇒ ΔS = –mL/T
= –(0.76kg) (333×103J/kg) (273K)
= –927 J/K
⇒ Change in entropy of the system during this process will be –927J/K.
The correct answer is: 927
Most Upvoted Answer
A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reve...
To solve this problem, we need to calculate the total heat required to melt the ice and then raise the temperature of the resulting liquid to 0°C.
1. The heat required to melt the ice can be calculated using the equation:
Q = m * Hf
where Q is the heat required, m is the mass of the ice, and Hf is the heat of fusion.
Given:
- Mass of ice = 262g = 0.262kg
- Heat of fusion for water = 334 J/g
Calculating the heat required to melt the ice:
Q = 0.262kg * 334 J/g = 87.308 J
2. Once the ice has melted, we need to raise the temperature of the liquid water to 0°C.
The specific heat capacity of water is 4.18 J/g°C.
Given:
- Mass of water = 1.78kg
Calculating the heat required to raise the temperature of the liquid water:
Q = m * c * ΔT
where Q is the heat required, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
ΔT = 0°C - (-273.15°C) = 273.15°C
Q = 1.78kg * 4.18 J/g°C * 273.15°C = 2037.69 J
3. Finally, we can calculate the total heat required by summing the heat required to melt the ice and the heat required to raise the temperature of the liquid water:
Total heat required = 87.308 J + 2037.69 J = 2124.998 J
Therefore, a total of approximately 2125 J of heat is required to melt the ice and raise the temperature of the resulting liquid water to 0°C.
1. The heat required to melt the ice can be calculated using the equation:
Q = m * Hf
where Q is the heat required, m is the mass of the ice, and Hf is the heat of fusion.
Given:
- Mass of ice = 262g = 0.262kg
- Heat of fusion for water = 334 J/g
Calculating the heat required to melt the ice:
Q = 0.262kg * 334 J/g = 87.308 J
2. Once the ice has melted, we need to raise the temperature of the liquid water to 0°C.
The specific heat capacity of water is 4.18 J/g°C.
Given:
- Mass of water = 1.78kg
Calculating the heat required to raise the temperature of the liquid water:
Q = m * c * ΔT
where Q is the heat required, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
ΔT = 0°C - (-273.15°C) = 273.15°C
Q = 1.78kg * 4.18 J/g°C * 273.15°C = 2037.69 J
3. Finally, we can calculate the total heat required by summing the heat required to melt the ice and the heat required to raise the temperature of the liquid water:
Total heat required = 87.308 J + 2037.69 J = 2124.998 J
Therefore, a total of approximately 2125 J of heat is required to melt the ice and raise the temperature of the resulting liquid water to 0°C.
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A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. Can you explain this answer?
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A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. 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 mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. 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 mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. Can you explain this answer?.
A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. 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 mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. 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 mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. Can you explain this answer?.
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A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. Can you explain this answer?, a detailed solution for A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. Can you explain this answer? has been provided alongside types of A mixture of 1.78kg of water and 262g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC. Calculate the entropy change of the system during this process (in J/K). L = 333 × 103 J/kg for waterCorrect answer is '927'. Can you explain this answer? theory, EduRev gives you an
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