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A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. 'm' grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of 'm' is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1 °C-1)
- a)2
- b)3.2
- c)2.6
- d)4
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A calorimeter of water equivalent 20 g contains 180 g of water at 25&d...
To calculate the final temperature of the water in the calorimeter, we can use the principle of conservation of energy.
The equation for conservation of energy is:
Energy gained by water = Energy lost by calorimeter
The energy gained by the water is given by the equation:
Energy gained = mass of water * specific heat capacity of water * change in temperature
The energy lost by the calorimeter is given by the equation:
Energy lost = water equivalent of calorimeter * specific heat capacity of water * change in temperature
Setting these two equations equal to each other, we can solve for the final temperature.
mass of water * specific heat capacity of water * change in temperature = water equivalent of calorimeter * specific heat capacity of water * change in temperature
180 g * 4.18 J/g°C * (final temperature - 25°C) = 20 g * 4.18 J/g°C * (final temperature - 25°C)
Simplifying the equation:
754.8 (final temperature - 25) = 83.6 (final temperature - 25)
Expanding the equation:
754.8 final temperature - 18870 = 83.6 final temperature - 2090
Combining like terms:
754.8 final temperature - 83.6 final temperature = 18870 - 2090
671.2 final temperature = 16780
Dividing both sides by 671.2:
final temperature = 16780 / 671.2
Final temperature ≈ 25 + 25
Therefore, the final temperature of the water in the calorimeter is approximately 50°C.
The equation for conservation of energy is:
Energy gained by water = Energy lost by calorimeter
The energy gained by the water is given by the equation:
Energy gained = mass of water * specific heat capacity of water * change in temperature
The energy lost by the calorimeter is given by the equation:
Energy lost = water equivalent of calorimeter * specific heat capacity of water * change in temperature
Setting these two equations equal to each other, we can solve for the final temperature.
mass of water * specific heat capacity of water * change in temperature = water equivalent of calorimeter * specific heat capacity of water * change in temperature
180 g * 4.18 J/g°C * (final temperature - 25°C) = 20 g * 4.18 J/g°C * (final temperature - 25°C)
Simplifying the equation:
754.8 (final temperature - 25) = 83.6 (final temperature - 25)
Expanding the equation:
754.8 final temperature - 18870 = 83.6 final temperature - 2090
Combining like terms:
754.8 final temperature - 83.6 final temperature = 18870 - 2090
671.2 final temperature = 16780
Dividing both sides by 671.2:
final temperature = 16780 / 671.2
Final temperature ≈ 25 + 25
Therefore, the final temperature of the water in the calorimeter is approximately 50°C.
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Community Answer
A calorimeter of water equivalent 20 g contains 180 g of water at 25&d...
Heat lost by steam = heat gained by water
180 × 1 × (31 - 25) + 20 × (31-25) = m × 540 + m × 1 × (100-31)
180 × 6 +20 × 6 = 540m + 100 m - 31m
1080 + 120 = 640 m - 31m
1200 = 609m
m = 1200/609
= 1.97
m = 2
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A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. m grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of m is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1°C-1)a)2b)3.2c)2.6d)4Correct answer is option 'A'. Can you explain this answer?
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A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. m grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of m is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1°C-1)a)2b)3.2c)2.6d)4Correct 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 of water equivalent 20 g contains 180 g of water at 25°C. m grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of m is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1°C-1)a)2b)3.2c)2.6d)4Correct 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 of water equivalent 20 g contains 180 g of water at 25°C. m grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of m is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1°C-1)a)2b)3.2c)2.6d)4Correct answer is option 'A'. Can you explain this answer?.
A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. m grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of m is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1°C-1)a)2b)3.2c)2.6d)4Correct 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 of water equivalent 20 g contains 180 g of water at 25°C. m grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of m is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1°C-1)a)2b)3.2c)2.6d)4Correct 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 of water equivalent 20 g contains 180 g of water at 25°C. m grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of m is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1°C-1)a)2b)3.2c)2.6d)4Correct answer is option 'A'. Can you explain this answer?.
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