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30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)
Correct answer is '0'. Can you explain this answer?
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30 gm of ice at 0°C is added to a beaker containing 60gm of water ...
The amount of heat required to melt the ice of mass 10gm at 0°C is
Q = ML = 30X 80 = 2 4 0 0 cal.
Where L is the latent heat of melting of ice and m is the mass of the ice. The amount of heat availabel in water of mass 60gm at 35°C is Q = MCvdT = 60 x 1 x 35 = 2100 cal. Since the heat available is less than the heat required to melt the ice therefore ice will not melt as a result the temperature of the system will be at 0°C.
Q = ML = 30X 80 = 2 4 0 0 cal.
Where L is the latent heat of melting of ice and m is the mass of the ice. The amount of heat availabel in water of mass 60gm at 35°C is Q = MCvdT = 60 x 1 x 35 = 2100 cal. Since the heat available is less than the heat required to melt the ice therefore ice will not melt as a result the temperature of the system will be at 0°C.
Most Upvoted Answer
30 gm of ice at 0°C is added to a beaker containing 60gm of water ...
°C is melted to form water at 0°C. The heat required to melt ice is calculated using the formula:
Q = m * ΔHf
Where:
Q = heat required
m = mass of ice
ΔHf = heat of fusion of ice
The heat of fusion of ice is 334 J/g.
Let's calculate the heat required:
Q = 30 g * 334 J/g
Q = 10,020 J
Therefore, 10,020 J of heat is required to melt 30 g of ice at 0°C.
Q = m * ΔHf
Where:
Q = heat required
m = mass of ice
ΔHf = heat of fusion of ice
The heat of fusion of ice is 334 J/g.
Let's calculate the heat required:
Q = 30 g * 334 J/g
Q = 10,020 J
Therefore, 10,020 J of heat is required to melt 30 g of ice at 0°C.
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30 gm of ice at 0°C is added to a beaker containing 60gm of water ...
The amount of heat required to melt the ice of mass 10gm at 0°C is
Q = ML = 30X 80 = 2 4 0 0 cal.
Where L is the latent heat of melting of ice and m is the mass of the ice. The amount of heat availabel in water of mass 60gm at 35°C is Q = MCvdT = 60 x 1 x 35 = 2100 cal. Since the heat available is less than the heat required to melt the ice therefore ice will not melt as a result the temperature of the system will be at 0°C.
Q = ML = 30X 80 = 2 4 0 0 cal.
Where L is the latent heat of melting of ice and m is the mass of the ice. The amount of heat availabel in water of mass 60gm at 35°C is Q = MCvdT = 60 x 1 x 35 = 2100 cal. Since the heat available is less than the heat required to melt the ice therefore ice will not melt as a result the temperature of the system will be at 0°C.
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30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. Can you explain this answer?
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30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. 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 30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. Can you explain this answer?.
30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. 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 30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. Can you explain this answer?.
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30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. Can you explain this answer?, a detailed solution for 30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. Can you explain this answer? has been provided alongside types of 30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)Correct answer is '0'. Can you explain this answer? theory, EduRev gives you an
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