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120 g of ice at 0°C is mixed with 100 g of water at 80°C. Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/g-°C. The final temperature of the mixture is
- a)0°C
- b)40°C
- c)20°C
- d)10°C
Correct answer is option 'A'. Can you explain this answer?
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120 g of ice at 0°C is mixed with 100 g of water at 80°C. Late...
°C is added to 200 g of water at 20°C.
To find the final temperature when the ice melts, we can use the principle of conservation of energy.
The heat gained by the water is equal to the heat lost by the ice.
The heat gained by the water is given by the equation:
Q = mcΔT
Where:
Q = heat gained
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
The heat lost by the ice is given by the equation:
Q = mLf
Where:
Q = heat lost
m = mass of ice
Lf = latent heat of fusion of ice
First, let's find the heat gained by the water:
Q_water = mcΔT
Q_water = 200 g * 4.18 J/g°C * (T - 20°C)
Next, let's find the heat lost by the ice:
Q_ice = mLf
Q_ice = 120 g * 334 J/g
Since the heat gained by the water is equal to the heat lost by the ice, we can set up the equation:
Q_water = Q_ice
200 g * 4.18 J/g°C * (T - 20°C) = 120 g * 334 J/g
Simplifying the equation:
836(T - 20) = 40080
Dividing both sides by 836:
T - 20 = 48
Adding 20 to both sides:
T = 68°C
Therefore, the final temperature when the ice melts is 68°C.
To find the final temperature when the ice melts, we can use the principle of conservation of energy.
The heat gained by the water is equal to the heat lost by the ice.
The heat gained by the water is given by the equation:
Q = mcΔT
Where:
Q = heat gained
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
The heat lost by the ice is given by the equation:
Q = mLf
Where:
Q = heat lost
m = mass of ice
Lf = latent heat of fusion of ice
First, let's find the heat gained by the water:
Q_water = mcΔT
Q_water = 200 g * 4.18 J/g°C * (T - 20°C)
Next, let's find the heat lost by the ice:
Q_ice = mLf
Q_ice = 120 g * 334 J/g
Since the heat gained by the water is equal to the heat lost by the ice, we can set up the equation:
Q_water = Q_ice
200 g * 4.18 J/g°C * (T - 20°C) = 120 g * 334 J/g
Simplifying the equation:
836(T - 20) = 40080
Dividing both sides by 836:
T - 20 = 48
Adding 20 to both sides:
T = 68°C
Therefore, the final temperature when the ice melts is 68°C.
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120 g of ice at 0°C is mixed with 100 g of water at 80°C. Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/g-°C. The final temperature of the mixture isa)0°Cb)40°Cc)20°Cd)10°CCorrect answer is option 'A'. Can you explain this answer?
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120 g of ice at 0°C is mixed with 100 g of water at 80°C. Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/g-°C. The final temperature of the mixture isa)0°Cb)40°Cc)20°Cd)10°CCorrect 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 120 g of ice at 0°C is mixed with 100 g of water at 80°C. Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/g-°C. The final temperature of the mixture isa)0°Cb)40°Cc)20°Cd)10°CCorrect 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 120 g of ice at 0°C is mixed with 100 g of water at 80°C. Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/g-°C. The final temperature of the mixture isa)0°Cb)40°Cc)20°Cd)10°CCorrect answer is option 'A'. Can you explain this answer?.
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