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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...
°C. A 10 g piece of metal at 100 °C is placed into the calorimeter. The specific heat capacity of the metal is 0.5 J/g°C. Determine 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 will be equal to the heat gained by the water and calorimeter. The formula to calculate heat is:
Q = m * c * ΔT
Where:
- Q is the heat exchanged
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature
First, let's calculate the heat lost by the metal. The initial temperature of the metal is 100 °C, and the final temperature is unknown.
Q_lost = m * c * ΔT
Q_lost = 10 g * 0.5 J/g°C * (100 °C - Tf)
Next, let's calculate the heat gained by the water and calorimeter. The initial temperature of the water is 50 °C, and the final temperature is also unknown.
Q_gained = m * c * ΔT
Q_gained = 50 g * 4.18 J/g°C * (Tf - 50 °C)
Since the principle of conservation of energy states that the heat lost is equal to the heat gained, we can set the two equations equal to each other:
10 g * 0.5 J/g°C * (100 °C - Tf) = 50 g * 4.18 J/g°C * (Tf - 50 °C)
Simplifying the equation:
5 J/°C * (100 °C - Tf) = 209 J/°C * (Tf - 50 °C)
500 °C - 5Tf = 209Tf - 10450 °C
214Tf = 10950 °C
Tf = 10950 °C / 214
Tf ≈ 51.16 °C
Therefore, the final temperature of the mixture is approximately 51.16 °C.
To solve this problem, we need to use the principle of conservation of energy. The heat lost by the metal will be equal to the heat gained by the water and calorimeter. The formula to calculate heat is:
Q = m * c * ΔT
Where:
- Q is the heat exchanged
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature
First, let's calculate the heat lost by the metal. The initial temperature of the metal is 100 °C, and the final temperature is unknown.
Q_lost = m * c * ΔT
Q_lost = 10 g * 0.5 J/g°C * (100 °C - Tf)
Next, let's calculate the heat gained by the water and calorimeter. The initial temperature of the water is 50 °C, and the final temperature is also unknown.
Q_gained = m * c * ΔT
Q_gained = 50 g * 4.18 J/g°C * (Tf - 50 °C)
Since the principle of conservation of energy states that the heat lost is equal to the heat gained, we can set the two equations equal to each other:
10 g * 0.5 J/g°C * (100 °C - Tf) = 50 g * 4.18 J/g°C * (Tf - 50 °C)
Simplifying the equation:
5 J/°C * (100 °C - Tf) = 209 J/°C * (Tf - 50 °C)
500 °C - 5Tf = 209Tf - 10450 °C
214Tf = 10950 °C
Tf = 10950 °C / 214
Tf ≈ 51.16 °C
Therefore, the final temperature of the mixture is approximately 51.16 °C.
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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?
Question Description
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 JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE 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 JEE 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?.
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 JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE 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 JEE 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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