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2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :
Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)
Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)
Correct answer is '6'. Can you explain this answer?
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2 kg ice at – 20°C is mixed with 5 kg water at 20°C. The...
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2 kg ice at – 20°C is mixed with 5 kg water at 20°C. The...
0 degrees Celsius is placed in a room at 25 degrees Celsius. Assuming no heat is lost to the surroundings, calculate the final temperature of the ice after it has melted.
To solve this problem, we need to calculate the heat gained by the ice and equate it to the heat lost by the room.
The heat gained by the ice can be calculated using the formula:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass of the ice (2 kg)
c = specific heat capacity of ice (2.09 J/g°C)
ΔT = change in temperature
Since the ice is melting, the change in temperature is from 0°C to the final temperature.
The heat lost by the room can be calculated using the formula:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass of the room (assumed to be 1 kg)
c = specific heat capacity of the room (assumed to be 1 J/g°C)
ΔT = change in temperature
Since the room is losing heat, the change in temperature is from 25°C to the final temperature.
Setting the two equations equal to each other, we have:
m1c1ΔT1 = m2c2ΔT2
(2 kg)(2.09 J/g°C)(ΔT1) = (1 kg)(1 J/g°C)(ΔT2)
Simplifying, we get:
(4.18 J/°C)(ΔT1) = (1 J/°C)(ΔT2)
Dividing both sides by 4.18 J/°C, we get:
ΔT1 = (1/4.18)(ΔT2)
Since the ice is melting, the final temperature (ΔT2) is 0°C. Substituting this value into the equation, we have:
ΔT1 = (1/4.18)(0)
ΔT1 = 0
This means that the final temperature of the ice after melting is 0°C, the same as its initial temperature.
To solve this problem, we need to calculate the heat gained by the ice and equate it to the heat lost by the room.
The heat gained by the ice can be calculated using the formula:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass of the ice (2 kg)
c = specific heat capacity of ice (2.09 J/g°C)
ΔT = change in temperature
Since the ice is melting, the change in temperature is from 0°C to the final temperature.
The heat lost by the room can be calculated using the formula:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass of the room (assumed to be 1 kg)
c = specific heat capacity of the room (assumed to be 1 J/g°C)
ΔT = change in temperature
Since the room is losing heat, the change in temperature is from 25°C to the final temperature.
Setting the two equations equal to each other, we have:
m1c1ΔT1 = m2c2ΔT2
(2 kg)(2.09 J/g°C)(ΔT1) = (1 kg)(1 J/g°C)(ΔT2)
Simplifying, we get:
(4.18 J/°C)(ΔT1) = (1 J/°C)(ΔT2)
Dividing both sides by 4.18 J/°C, we get:
ΔT1 = (1/4.18)(ΔT2)
Since the ice is melting, the final temperature (ΔT2) is 0°C. Substituting this value into the equation, we have:
ΔT1 = (1/4.18)(0)
ΔT1 = 0
This means that the final temperature of the ice after melting is 0°C, the same as its initial temperature.
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2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer?
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2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer?.
2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer?.
Solutions for 2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer?, a detailed solution for 2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer? has been provided alongside types of 2 kg ice at – 20°C is mixed with 5 kg water at 20°C. Then final amount of water in the mixture would be :Given specific heat of ice = 0.5cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion of ice = 80 cal/g. (in kg)Correct answer is '6'. Can you explain this answer? theory, EduRev gives you an
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