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4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.
- a)18 gm
- b)20 gm
- c)22 gm
- d)24 gm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
4 gms of steam at 100°C is added to 20 gms of water at 46°C in...
Heat released by steam inconversion to water at 100°C is Q1 = mL = 4 × 540 = 2160 cal.
Heat required to raise temperature of water from 46°C t 100°C is Q2 = mS Δθ = 20 × 1 × 54 = 1080 J
Hence all steam is not converted to water only half steam shall be converted to water
So Final mass of water = 20 + 2 = 22 gm
Heat required to raise temperature of water from 46°C t 100°C is Q2 = mS Δθ = 20 × 1 × 54 = 1080 J
Hence all steam is not converted to water only half steam shall be converted to water
So Final mass of water = 20 + 2 = 22 gm
Most Upvoted Answer
4 gms of steam at 100°C is added to 20 gms of water at 46°C in...
Concept Overview
To find the thermal equilibrium of the system, we need to consider the heat lost by the steam as it condenses and cools down, and the heat gained by the water as it warms up.
Step 1: Heat Lost by Steam
- The steam will first condense into water at 100°C, releasing latent heat.
- Mass of steam = 4 g
- Latent heat of vaporization = 540 cal/g
- Heat released by 4 g of steam = 4 g * 540 cal/g = 2160 cal
After condensing, this water will cool down from 100°C to the final equilibrium temperature (T).
Step 2: Heat Lost by Condensed Steam
- Heat lost by this water cooling down:
- Mass of condensed steam = 4 g
- Specific heat of water = 1 cal/g°C
- Heat lost = mass * specific heat * change in temperature = 4 g * 1 cal/g°C * (100 - T) °C
Total heat lost by steam = 2160 cal + 4 * (100 - T)
Step 3: Heat Gained by Water
- The initial water at 46°C will gain heat as it warms up to T.
- Mass of water = 20 g
- Change in temperature = (T - 46) °C
- Heat gained = mass * specific heat * change in temperature = 20 g * 1 cal/g°C * (T - 46) °C
Step 4: Setting up the Equation
Now, we set the heat lost equal to the heat gained:
2160 + 4 * (100 - T) = 20 * (T - 46)
Step 5: Solving the Equation
After simplifying and solving the equation, you'll find that T approaches approximately 63°C.
Now, plug T back into the heat equations to find the total mass of water at thermal equilibrium:
- Total mass of water = 20 g (initial water) + 4 g (condensed steam) = 24 g.
Conclusion
Thus, the mass of water in the container at thermal equilibrium is 24 grams (option C).
To find the thermal equilibrium of the system, we need to consider the heat lost by the steam as it condenses and cools down, and the heat gained by the water as it warms up.
Step 1: Heat Lost by Steam
- The steam will first condense into water at 100°C, releasing latent heat.
- Mass of steam = 4 g
- Latent heat of vaporization = 540 cal/g
- Heat released by 4 g of steam = 4 g * 540 cal/g = 2160 cal
After condensing, this water will cool down from 100°C to the final equilibrium temperature (T).
Step 2: Heat Lost by Condensed Steam
- Heat lost by this water cooling down:
- Mass of condensed steam = 4 g
- Specific heat of water = 1 cal/g°C
- Heat lost = mass * specific heat * change in temperature = 4 g * 1 cal/g°C * (100 - T) °C
Total heat lost by steam = 2160 cal + 4 * (100 - T)
Step 3: Heat Gained by Water
- The initial water at 46°C will gain heat as it warms up to T.
- Mass of water = 20 g
- Change in temperature = (T - 46) °C
- Heat gained = mass * specific heat * change in temperature = 20 g * 1 cal/g°C * (T - 46) °C
Step 4: Setting up the Equation
Now, we set the heat lost equal to the heat gained:
2160 + 4 * (100 - T) = 20 * (T - 46)
Step 5: Solving the Equation
After simplifying and solving the equation, you'll find that T approaches approximately 63°C.
Now, plug T back into the heat equations to find the total mass of water at thermal equilibrium:
- Total mass of water = 20 g (initial water) + 4 g (condensed steam) = 24 g.
Conclusion
Thus, the mass of water in the container at thermal equilibrium is 24 grams (option C).
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4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.a)18 gmb)20 gmc)22 gmd)24 gmCorrect answer is option 'C'. Can you explain this answer?
Question Description
4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.a)18 gmb)20 gmc)22 gmd)24 gmCorrect answer is option 'C'. 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 4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.a)18 gmb)20 gmc)22 gmd)24 gmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.a)18 gmb)20 gmc)22 gmd)24 gmCorrect answer is option 'C'. Can you explain this answer?.
4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.a)18 gmb)20 gmc)22 gmd)24 gmCorrect answer is option 'C'. 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 4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.a)18 gmb)20 gmc)22 gmd)24 gmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.a)18 gmb)20 gmc)22 gmd)24 gmCorrect answer is option 'C'. Can you explain this answer?.
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