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30g water at 30°C is mixed with 1g of steam at 100℃. Calculate the final temperature of the mixture. Sw = 1cal/ g°C, Lvap= 540cal/g.
- a)49.67°C
- b)38.33°C
- c)53.67°C
- d)41.33°C
Correct answer is option 'A'. Can you explain this answer?
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30g water at 30°C is mixed with 1g of steam at 100. Calculate the ...
°C is added to 50g of water at 60°C. What is the final temperature of the mixture?
To solve this problem, we need to use the principle of conservation of energy. The total amount of energy in the system (water) remains constant.
The formula to calculate the final temperature of the mixture is:
m1c1ΔT1 + m2c2ΔT2 = 0
Where:
m1 = mass of water 1
c1 = specific heat of water 1
ΔT1 = change in temperature of water 1
m2 = mass of water 2
c2 = specific heat of water 2
ΔT2 = change in temperature of water 2
Let's substitute the values given in the problem:
m1 = 30g
c1 = 1 cal/g°C (specific heat of water)
ΔT1 = Tf - 30°C (final temperature - initial temperature)
m2 = 50g
c2 = 1 cal/g°C (specific heat of water)
ΔT2 = Tf - 60°C (final temperature - initial temperature)
The formula becomes:
30g × 1 cal/g°C × (Tf - 30°C) + 50g × 1 cal/g°C × (Tf - 60°C) = 0
Simplifying:
30(Tf - 30) + 50(Tf - 60) = 0
30Tf - 900 + 50Tf - 3000 = 0
80Tf - 3900 = 0
80Tf = 3900
Tf = 48.75°C
Therefore, the final temperature of the mixture is 48.75°C.
To solve this problem, we need to use the principle of conservation of energy. The total amount of energy in the system (water) remains constant.
The formula to calculate the final temperature of the mixture is:
m1c1ΔT1 + m2c2ΔT2 = 0
Where:
m1 = mass of water 1
c1 = specific heat of water 1
ΔT1 = change in temperature of water 1
m2 = mass of water 2
c2 = specific heat of water 2
ΔT2 = change in temperature of water 2
Let's substitute the values given in the problem:
m1 = 30g
c1 = 1 cal/g°C (specific heat of water)
ΔT1 = Tf - 30°C (final temperature - initial temperature)
m2 = 50g
c2 = 1 cal/g°C (specific heat of water)
ΔT2 = Tf - 60°C (final temperature - initial temperature)
The formula becomes:
30g × 1 cal/g°C × (Tf - 30°C) + 50g × 1 cal/g°C × (Tf - 60°C) = 0
Simplifying:
30(Tf - 30) + 50(Tf - 60) = 0
30Tf - 900 + 50Tf - 3000 = 0
80Tf - 3900 = 0
80Tf = 3900
Tf = 48.75°C
Therefore, the final temperature of the mixture is 48.75°C.
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Community Answer
30g water at 30°C is mixed with 1g of steam at 100. Calculate the ...
30*1*(T-30) = 1*540 + 1*1*(100-T)
∴ 30T – 900 = 540 + 100 – T
⇒ 31T = 1540
⇒ T = 49.67°C
∴ 30T – 900 = 540 + 100 – T
⇒ 31T = 1540
⇒ T = 49.67°C
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30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect answer is option 'A'. Can you explain this answer?
Question Description
30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect 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 30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect 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 30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect answer is option 'A'. Can you explain this answer?.
30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect 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 30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect 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 30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect answer is option 'A'. Can you explain this answer?.
Solutions for 30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for 30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of 30g water at 30°C is mixed with 1g of steam at 100. Calculate the final temperature of the mixture. Sw= 1cal/ g°C, Lvap= 540cal/g.a)49.67°Cb)38.33°Cc)53.67°Cd)41.33°CCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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