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If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then result it temperature of the mixture is?
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If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then resu...
Given:
- Mass of steam = 1g
- Initial temperature of steam = 100°C
- Mass of ice = 1g
- Initial temperature of ice = 0°C
To find:
- The final temperature of the mixture
Assumptions:
- The system is isolated, meaning no heat is lost or gained to the surroundings.
- The specific heat capacities of steam and ice are constant throughout the process.
- There is no phase change during the process.
Solution:
1. Calculate the heat gained by the ice to reach its melting point:
- The heat gained by the ice can be calculated using the formula: Q = mcΔT, where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- The specific heat capacity of ice is 2.09 J/g°C.
- The change in temperature is 0 - (-10) = 10°C (since the melting point of ice is 0°C).
- Therefore, the heat gained by the ice is Q = (1g)(2.09 J/g°C)(10°C) = 20.9 J.
2. Calculate the heat lost by the steam to reach its condensation point:
- The heat lost by the steam can also be calculated using the formula: Q = mcΔT.
- The specific heat capacity of steam is 2.03 J/g°C.
- The change in temperature is 100 - 100 = 0°C (since the condensation point of steam is 100°C).
- Therefore, the heat lost by the steam is Q = (1g)(2.03 J/g°C)(0°C) = 0 J.
3. Calculate the heat released during the phase change of the steam:
- The heat released during the phase change of the steam is given by the formula: Q = mL, where m is the mass and L is the latent heat of vaporization.
- The latent heat of vaporization of water is 2260 J/g.
- The mass of steam is 1g.
- Therefore, the heat released during the phase change is Q = (1g)(2260 J/g) = 2260 J.
4. Calculate the final temperature of the mixture:
- The total heat gained by the ice and the total heat lost by the steam should be equal.
- Therefore, 20.9 J + 0 J + 2260 J = 0 J + 20.9 J + 2260 J.
- This implies that the heat gained by the ice is equal to the heat released during the phase change of the steam.
- Since the specific heat capacity of water is 4.18 J/g°C, the heat gained by the ice can raise its temperature by Q/(mc) = 20.9 J/(1g)(4.18 J/g°C) = 5°C.
- Therefore, the final temperature of the mixture is 0°C + 5°C = 5°C.
Conclusion:
The final temperature of the mixture of 1g of steam at 100°C and 1g of ice at 0°C is 5°C.
- Mass of steam = 1g
- Initial temperature of steam = 100°C
- Mass of ice = 1g
- Initial temperature of ice = 0°C
To find:
- The final temperature of the mixture
Assumptions:
- The system is isolated, meaning no heat is lost or gained to the surroundings.
- The specific heat capacities of steam and ice are constant throughout the process.
- There is no phase change during the process.
Solution:
1. Calculate the heat gained by the ice to reach its melting point:
- The heat gained by the ice can be calculated using the formula: Q = mcΔT, where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- The specific heat capacity of ice is 2.09 J/g°C.
- The change in temperature is 0 - (-10) = 10°C (since the melting point of ice is 0°C).
- Therefore, the heat gained by the ice is Q = (1g)(2.09 J/g°C)(10°C) = 20.9 J.
2. Calculate the heat lost by the steam to reach its condensation point:
- The heat lost by the steam can also be calculated using the formula: Q = mcΔT.
- The specific heat capacity of steam is 2.03 J/g°C.
- The change in temperature is 100 - 100 = 0°C (since the condensation point of steam is 100°C).
- Therefore, the heat lost by the steam is Q = (1g)(2.03 J/g°C)(0°C) = 0 J.
3. Calculate the heat released during the phase change of the steam:
- The heat released during the phase change of the steam is given by the formula: Q = mL, where m is the mass and L is the latent heat of vaporization.
- The latent heat of vaporization of water is 2260 J/g.
- The mass of steam is 1g.
- Therefore, the heat released during the phase change is Q = (1g)(2260 J/g) = 2260 J.
4. Calculate the final temperature of the mixture:
- The total heat gained by the ice and the total heat lost by the steam should be equal.
- Therefore, 20.9 J + 0 J + 2260 J = 0 J + 20.9 J + 2260 J.
- This implies that the heat gained by the ice is equal to the heat released during the phase change of the steam.
- Since the specific heat capacity of water is 4.18 J/g°C, the heat gained by the ice can raise its temperature by Q/(mc) = 20.9 J/(1g)(4.18 J/g°C) = 5°C.
- Therefore, the final temperature of the mixture is 0°C + 5°C = 5°C.
Conclusion:
The final temperature of the mixture of 1g of steam at 100°C and 1g of ice at 0°C is 5°C.
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If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then result it temperature of the mixture is?
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If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then result it temperature of the mixture is? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then result it temperature of the mixture is? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then result it temperature of the mixture is?.
If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then result it temperature of the mixture is? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then result it temperature of the mixture is? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 1g of steam at 100 C steam is mixed with 1g of ice at 0 C then result it temperature of the mixture is?.
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