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2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]
- a)6 kg
- b)7 kg
- c)3.5 kg
- d)5 kg
Correct answer is option 'A'. Can you explain this answer?
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2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then...
To find the final temperature of the ice, we need to know the specific heat capacity of the ice. Let's assume the specific heat capacity of ice is 2.09 J/g°C.
First, we need to convert 2 kg of ice into grams. There are 1000 grams in 1 kg, so 2 kg is equal to 2000 grams.
Next, we can use the formula: Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since the ice is at -20°C and we want to find the final temperature, we can assume the change in temperature is ΔT = Tf - (-20).
Let's assume the final temperature is Tf. The heat energy gained by the ice can be calculated as:
Q = 2000 g * 2.09 J/g°C * (Tf - (-20))
Since the ice is gaining heat energy, the Q value must be positive. Let's solve for Tf:
2000 g * 2.09 J/g°C * (Tf - (-20)) > 0
Simplifying the equation:
2000 g * 2.09 J/g°C * (Tf + 20) > 0
(2000 * 2.09) * (Tf + 20) > 0
4180 * (Tf + 20) > 0
Tf + 20 > 0
Tf > -20
So, the final temperature Tf must be greater than -20°C for the ice to gain heat energy.
First, we need to convert 2 kg of ice into grams. There are 1000 grams in 1 kg, so 2 kg is equal to 2000 grams.
Next, we can use the formula: Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since the ice is at -20°C and we want to find the final temperature, we can assume the change in temperature is ΔT = Tf - (-20).
Let's assume the final temperature is Tf. The heat energy gained by the ice can be calculated as:
Q = 2000 g * 2.09 J/g°C * (Tf - (-20))
Since the ice is gaining heat energy, the Q value must be positive. Let's solve for Tf:
2000 g * 2.09 J/g°C * (Tf - (-20)) > 0
Simplifying the equation:
2000 g * 2.09 J/g°C * (Tf + 20) > 0
(2000 * 2.09) * (Tf + 20) > 0
4180 * (Tf + 20) > 0
Tf + 20 > 0
Tf > -20
So, the final temperature Tf must be greater than -20°C for the ice to gain heat energy.
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2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer?
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2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer?.
2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer?.
Solutions for 2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for 2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of 2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 2 kg ice at - 20 ºC is mixed with 5 kg water at 20 ºC . Then final amount of water in the mixture would be : [Given : specific heat of ice = 0.5 cal/g ºC, specific heat of water = 1 cal/g ºC, latent heat of fusion of ice = 80 cal/gm]a)6 kgb)7 kgc)3.5 kgd)5 kgCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice NEET tests.
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