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If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is [1999]
- a)270ºC
- b)230ºC
- c)100ºC
- d)50ºC
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If 1 g of steam is mixed with 1 g of ice, then the resultant temperatu...
Heat required by ice a t 0° C to reach a temperature of 100°C = mL + mcΔθ = 1 × 80 + 1 × 1 × (100 – 0) = 180 cal
Heat available with 1 g steam to condense into 1 g of water at 100°C = 536 cal.
Obviously the whole steam will not be condensed and ice will attain a temperature of 100°C; so the temperature of mixture = 100°C.
Heat available with 1 g steam to condense into 1 g of water at 100°C = 536 cal.
Obviously the whole steam will not be condensed and ice will attain a temperature of 100°C; so the temperature of mixture = 100°C.
Most Upvoted Answer
If 1 g of steam is mixed with 1 g of ice, then the resultant temperatu...
The process of mixing steam and ice is an example of a phase change. When steam (water vapor) condenses to form ice, it releases a certain amount of heat energy. This heat energy is used to melt the ice, so that both substances end up at the same temperature.
The key concept to understand here is the heat of fusion, which is the amount of heat energy required to melt one gram of ice at its melting point (0°C). The heat of fusion for water is 334 J/g.
So, when 1 g of steam is mixed with 1 g of ice, the steam condenses and releases 2260 J of heat energy (the heat of condensation for water is 2260 J/g). This heat energy is used to melt the ice, which requires 334 J of heat energy. The remaining heat energy (2260 - 334 = 1926 J) is used to raise the temperature of the resulting water to its final temperature.
To calculate the final temperature, we can use the formula:
Q = mCΔT
Where Q is the heat energy transferred, m is the mass of the water, C is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature.
We know that Q = 1926 J, m = 2 g (the total mass of the water), and C = 4.18 J/g°C. Solving for ΔT, we get:
ΔT = Q / (mC) = 1926 / (2 x 4.18) = 230.6°C
Therefore, the final temperature of the mixture is:
0°C + 230.6°C = 230.6°C
Answer: b) 230.6°C
The key concept to understand here is the heat of fusion, which is the amount of heat energy required to melt one gram of ice at its melting point (0°C). The heat of fusion for water is 334 J/g.
So, when 1 g of steam is mixed with 1 g of ice, the steam condenses and releases 2260 J of heat energy (the heat of condensation for water is 2260 J/g). This heat energy is used to melt the ice, which requires 334 J of heat energy. The remaining heat energy (2260 - 334 = 1926 J) is used to raise the temperature of the resulting water to its final temperature.
To calculate the final temperature, we can use the formula:
Q = mCΔT
Where Q is the heat energy transferred, m is the mass of the water, C is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature.
We know that Q = 1926 J, m = 2 g (the total mass of the water), and C = 4.18 J/g°C. Solving for ΔT, we get:
ΔT = Q / (mC) = 1926 / (2 x 4.18) = 230.6°C
Therefore, the final temperature of the mixture is:
0°C + 230.6°C = 230.6°C
Answer: b) 230.6°C
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Community Answer
If 1 g of steam is mixed with 1 g of ice, then the resultant temperatu...
Heat required by 1 g ice at 0C to melt into 1 g water at 0C is
Q1= mL (Latent heat of fusion 80 ca l/g)
=1×80 =80 cal.
Heat required by 1g of water at 0C to boil at100C is Q2=mc(specific heat of water c = 1cal/gC) ∆T =
1×1(100-0) =100 cal.
Thus, total heat required by 1g of ice to reach a temperature of100C is
Q =Q1+Q2= 80 + 100 =180 cal.
Heat available with 1 g of ice to condense into 1g of water at100C is
Q'=mL'(Latent of vaporisation= 536 cal/g)=1×536 cal = 536 cal .
Obviously, the whole steam will not be condensed and ice will attain temperature of100C. Thus, the temperature of mixture is100C, i e. op D.
Q1= mL (Latent heat of fusion 80 ca l/g)
=1×80 =80 cal.
Heat required by 1g of water at 0C to boil at100C is Q2=mc(specific heat of water c = 1cal/gC) ∆T =
1×1(100-0) =100 cal.
Thus, total heat required by 1g of ice to reach a temperature of100C is
Q =Q1+Q2= 80 + 100 =180 cal.
Heat available with 1 g of ice to condense into 1g of water at100C is
Q'=mL'(Latent of vaporisation= 536 cal/g)=1×536 cal = 536 cal .
Obviously, the whole steam will not be condensed and ice will attain temperature of100C. Thus, the temperature of mixture is100C, i e. op D.
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If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is [1999]a)270ºCb)230ºCc)100ºCd)50ºCCorrect answer is option 'C'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is [1999]a)270ºCb)230ºCc)100ºCd)50ºCCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is [1999]a)270ºCb)230ºCc)100ºCd)50ºCCorrect answer is option 'C'. Can you explain this answer?.
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