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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)4
- c)3.2
- d)2.6
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A calorimeter of water equivalent 20 g contains 180 g of water at 25°...
Heat given by water = mwCw (Tmix-Tw)
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= 200 x 1x (31- 25)
Heat taken by steam = m Lstem + m Cw (Ts – Tmix)
= m × 540 + m (1) × (100 –31)
= m × 540 + m (1) × (69)
From the principal of calorimeter,
Heat lost = Heat gained
∴ (200)(31 - 25) = m x 540+m(1)(69)
⇒ 1200 = m(609) ⇒m≈ 2.
Most Upvoted Answer
A calorimeter of water equivalent 20 g contains 180 g of water at 25°...
Given Information:
- Water equivalent: 20 g
- Temperature of water in calorimeter: 25°C
- Mass of water in calorimeter: 180 g
- Temperature of steam: 100°C
- Final temperature of mixture: 31°C
- Latent heat of water: 540 cal g–1
- Specific heat of water: 1 cal g–1 °C–1
Approach:
To solve this problem, we can use the principle of conservation of heat. The heat gained by the water and calorimeter must be equal to the heat lost by the steam.
Calculation:
Let's break down the calculation into steps:
1. Heat lost by the steam = Heat gained by the water and calorimeter
2. Heat lost by the steam = m * latent heat of water (as steam condenses to water)
3. Heat gained by the water and calorimeter = (mass of water + water equivalent) * specific heat of water * change in temperature
4. Substitute the given values into the equation:
m * 540 cal/g = (180 g + 20 g) * 1 cal/g°C * (31°C - 25°C)
5. Simplify the equation:
m * 540 cal/g = 200 g * 6 cal/g
6. Cancel out the units:
m * 540 = 1200
7. Solve for m:
m = 1200 / 540
m ≈ 2.22
Since we need to choose the closest option, the value of m is close to 2. Therefore, the correct answer is option A) 2.
- Water equivalent: 20 g
- Temperature of water in calorimeter: 25°C
- Mass of water in calorimeter: 180 g
- Temperature of steam: 100°C
- Final temperature of mixture: 31°C
- Latent heat of water: 540 cal g–1
- Specific heat of water: 1 cal g–1 °C–1
Approach:
To solve this problem, we can use the principle of conservation of heat. The heat gained by the water and calorimeter must be equal to the heat lost by the steam.
Calculation:
Let's break down the calculation into steps:
1. Heat lost by the steam = Heat gained by the water and calorimeter
2. Heat lost by the steam = m * latent heat of water (as steam condenses to water)
3. Heat gained by the water and calorimeter = (mass of water + water equivalent) * specific heat of water * change in temperature
4. Substitute the given values into the equation:
m * 540 cal/g = (180 g + 20 g) * 1 cal/g°C * (31°C - 25°C)
5. Simplify the equation:
m * 540 cal/g = 200 g * 6 cal/g
6. Cancel out the units:
m * 540 = 1200
7. Solve for m:
m = 1200 / 540
m ≈ 2.22
Since we need to choose the closest option, the value of m is close to 2. Therefore, the correct answer is option A) 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)4c)3.2d)2.6Correct 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)4c)3.2d)2.6Correct 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 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)4c)3.2d)2.6Correct 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 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)4c)3.2d)2.6Correct 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)4c)3.2d)2.6Correct 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 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)4c)3.2d)2.6Correct 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 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)4c)3.2d)2.6Correct answer is option 'A'. Can you explain this answer?.
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