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Steam at 100 degreeC is passed into 20g of water at 10 degree C, when water acquires a temperature of 80 degree C, the mass of water present will be? ( s of water 1, L of steam 540)?
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Steam at 100 degreeC is passed into 20g of water at 10 degree C, when ...
Given:
- Initial temperature of water (T1) = 10°C
- Final temperature of water (T2) = 80°C
- Mass of water (m1) = 20g
- Specific heat capacity of water (s) = 1 cal/g°C
- Latent heat of vaporization of water (L) = 540 cal/g
Approach:
To solve this problem, we will use the principle of heat transfer. The heat gained by the water is equal to the heat lost by the steam. We can calculate the heat gained by the water using the formula:
Q = m1 * s * (T2 - T1)
Where:
- Q is the heat gained by the water
- m1 is the mass of water
- s is the specific heat capacity of water
- T2 is the final temperature of water
- T1 is the initial temperature of water
We can then equate this heat gained by the water to the heat lost by the steam, which is given by:
Q = m2 * L
Where:
- m2 is the mass of steam
By equating these two expressions for Q, we can solve for m2, which will give us the mass of water present at the final temperature.
Calculation:
Step 1: Calculate the heat gained by the water
Q = m1 * s * (T2 - T1)
= 20g * 1 cal/g°C * (80°C - 10°C)
= 20g * 1 cal/g°C * 70°C
= 1400 cal
Step 2: Equate the heat gained by the water to the heat lost by the steam
Q = m2 * L
1400 cal = m2 * 540 cal/g
Step 3: Solve for m2 (mass of steam)
m2 = 1400 cal / 540 cal/g
≈ 2.59 g
Therefore, the mass of water present at the final temperature of 80°C is approximately 2.59 g.
- Initial temperature of water (T1) = 10°C
- Final temperature of water (T2) = 80°C
- Mass of water (m1) = 20g
- Specific heat capacity of water (s) = 1 cal/g°C
- Latent heat of vaporization of water (L) = 540 cal/g
Approach:
To solve this problem, we will use the principle of heat transfer. The heat gained by the water is equal to the heat lost by the steam. We can calculate the heat gained by the water using the formula:
Q = m1 * s * (T2 - T1)
Where:
- Q is the heat gained by the water
- m1 is the mass of water
- s is the specific heat capacity of water
- T2 is the final temperature of water
- T1 is the initial temperature of water
We can then equate this heat gained by the water to the heat lost by the steam, which is given by:
Q = m2 * L
Where:
- m2 is the mass of steam
By equating these two expressions for Q, we can solve for m2, which will give us the mass of water present at the final temperature.
Calculation:
Step 1: Calculate the heat gained by the water
Q = m1 * s * (T2 - T1)
= 20g * 1 cal/g°C * (80°C - 10°C)
= 20g * 1 cal/g°C * 70°C
= 1400 cal
Step 2: Equate the heat gained by the water to the heat lost by the steam
Q = m2 * L
1400 cal = m2 * 540 cal/g
Step 3: Solve for m2 (mass of steam)
m2 = 1400 cal / 540 cal/g
≈ 2.59 g
Therefore, the mass of water present at the final temperature of 80°C is approximately 2.59 g.
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Steam at 100 degreeC is passed into 20g of water at 10 degree C, when water acquires a temperature of 80 degree C, the mass of water present will be? ( s of water 1, L of steam 540)?
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Steam at 100 degreeC is passed into 20g of water at 10 degree C, when water acquires a temperature of 80 degree C, the mass of water present will be? ( s of water 1, L of steam 540)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Steam at 100 degreeC is passed into 20g of water at 10 degree C, when water acquires a temperature of 80 degree C, the mass of water present will be? ( s of water 1, L of steam 540)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Steam at 100 degreeC is passed into 20g of water at 10 degree C, when water acquires a temperature of 80 degree C, the mass of water present will be? ( s of water 1, L of steam 540)?.
Steam at 100 degreeC is passed into 20g of water at 10 degree C, when water acquires a temperature of 80 degree C, the mass of water present will be? ( s of water 1, L of steam 540)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Steam at 100 degreeC is passed into 20g of water at 10 degree C, when water acquires a temperature of 80 degree C, the mass of water present will be? ( s of water 1, L of steam 540)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Steam at 100 degreeC is passed into 20g of water at 10 degree C, when water acquires a temperature of 80 degree C, the mass of water present will be? ( s of water 1, L of steam 540)?.
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