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1 g of ice at 0C is mixed with 1 g of steam at 100∘C. After thermal equilibrium is achieved, the temperature of the mixture is?
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1 g of ice at 0C is mixed with 1 g of steam at 100∘C. After thermal eq...
Introduction:
When ice at 0°C is mixed with steam at 100°C, heat exchange occurs between the two substances until they reach thermal equilibrium. This process involves the transfer of heat energy from the steam to the ice, resulting in a change in temperature and phase of both substances. Let's analyze this process in detail.

Phase Changes:
When the ice and steam are mixed, the following phase changes occur:

1. Melting of ice: The ice at 0°C absorbs heat energy from the steam, causing it to melt and change from a solid to a liquid phase. During this process, the temperature of the ice remains constant at 0°C until all the ice has melted.

2. Vaporization of steam: The steam at 100°C releases heat energy to the ice, causing it to condense and change from a gaseous to a liquid phase. The temperature of the steam also remains constant at 100°C until all the steam has condensed.

Heat Transfer:
The heat transfer between the ice and steam can be calculated using the equation:

Q = mcΔT

Where:
- Q is the heat transferred
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature

In this case, the heat transferred from the steam to the ice is equal to the heat gained by the ice. Since the final temperature of the mixture is not given, we need to consider two scenarios:

1. If the final temperature is below 0°C:
- The heat transferred from the steam causes the ice to melt, resulting in an increase in temperature until the mixture reaches 0°C.
- The remaining heat is then used to raise the temperature of the water from 0°C to the final temperature.

2. If the final temperature is above 0°C:
- The heat transferred from the steam causes the ice to melt, resulting in an increase in temperature until the mixture reaches 0°C.
- The remaining heat is then used to raise the temperature of the water from 0°C to the final temperature.

Conclusion:
In conclusion, when 1 g of ice at 0°C is mixed with 1 g of steam at 100°C, the final temperature of the mixture will depend on the amount of heat transferred and the specific heat capacities of the ice and water. The specific heat capacity of ice is 2.09 J/g°C, and the specific heat capacity of water is 4.18 J/g°C. Therefore, by calculating the heat transfer and considering the specific heat capacities, we can determine the final temperature of the mixture.
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1 g of ice at 0C is mixed with 1 g of steam at 100∘C. After thermal equilibrium is achieved, the temperature of the mixture is?
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1 g of ice at 0C is mixed with 1 g of steam at 100∘C. After thermal equilibrium is achieved, the 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 1 g of ice at 0C is mixed with 1 g of steam at 100∘C. After thermal equilibrium is achieved, the temperature of the mixture is? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 g of ice at 0C is mixed with 1 g of steam at 100∘C. After thermal equilibrium is achieved, the temperature of the mixture is?.
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