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A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. then  the water equivalent of the calorimeter is :
  • a)
    12.5 g
  • b)
    6.25 g
  • c)
    25 g    
  • d)
    15 g
Correct answer is option 'A'. Can you explain this answer?
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A calorimeter contains 50 g of water at 50°C. The temperature fall...
°F. A 10 g piece of metal at 150°C is dropped into the calorimeter. Assuming no heat is lost to the surroundings, what is the final temperature of the mixture?

To solve this problem, we need to use the principle of conservation of energy. The heat lost by the metal is equal to the heat gained by the water and the calorimeter.

The heat lost by the metal can be calculated using the formula:

Q = mcΔT

where Q is the heat lost, m is the mass of the metal, c is the specific heat capacity of the metal, and ΔT is the change in temperature.

Assuming the metal is made of aluminum, which has a specific heat of 0.902 J/g°C, we can calculate the heat lost by the metal:

Q = 10 g × 0.902 J/g°C × (150°C - Tf)

where Tf is the final temperature of the mixture.

The heat gained by the water can be calculated using the formula:

Q = mcΔT

where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Using the values given in the problem, we can calculate the heat gained by the water:

Q = 50 g × 4.184 J/g°C × (Tf - 50°C)

The heat gained by the calorimeter is assumed to be negligible since its mass is much smaller than that of the water and the metal.

Since the total heat lost by the metal is equal to the total heat gained by the water and the calorimeter, we can set the two equations equal to each other and solve for Tf:

10 g × 0.902 J/g°C × (150°C - Tf) = 50 g × 4.184 J/g°C × (Tf - 50°C)

Simplifying the equation, we get:

1351.8 - 9.02Tf = 209.2Tf - 10460

Combining like terms, we get:

210.22Tf = 11812

Dividing both sides by 210.22, we get:

Tf ≈ 56.1°C

Therefore, the final temperature of the mixture is approximately 56.1°C.
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A calorimeter contains 50 g of water at 50°C. The temperature fall...
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A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. then the water equivalent of the calorimeter is :a)12.5 gb)6.25 gc)25 g d)15 gCorrect answer is option 'A'. Can you explain this answer?
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A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. then the water equivalent of the calorimeter is :a)12.5 gb)6.25 gc)25 g d)15 gCorrect 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 A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. then the water equivalent of the calorimeter is :a)12.5 gb)6.25 gc)25 g d)15 gCorrect 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 A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. then the water equivalent of the calorimeter is :a)12.5 gb)6.25 gc)25 g d)15 gCorrect answer is option 'A'. Can you explain this answer?.
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