Can Adding Ice and Steam Change Water Temperature?

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In summary, the conversation involves solving for the equilibrium temperature and original mass of a closed system containing boiling water, ice, and steam. For the first question, Judy must add a certain amount of -12.0 C ice to the thermos in order to reach an equilibrium temperature of 75.0 C. For the second question, the equation (Mice)(Lf) + (Mw)(Cw)(Change in Temp) = (Ms)(Lv) + (Msw)(Cw)(Change in Temp) can be used to solve for the original mass of steam at 100 C. Finally, for the third question, the equation (Mass of water)(Specific heat capacity of water)(Temp of water
  • #1
jayphysics
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1.) Judy places .150 kg of boiling water in a thermos bottle. How many kilograms of ice at -12.0 C must Judy add to the thermos so that the equilbrium temperature of the water is 75.0 C?

2.) A 0.03 kg ice cube at 0C is placed in an insulated box that contains a fixed quantity of steam at 100 C. When thermal equilibrium of this closed system is established, its temperature is found to be 23C. Determine the original mass of the steam at 100 C.

3.) What will the final temperature if 1 kg of water at 20C is heated by the condensation of 30 g steam at 100 C?
 
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Hi jayphysics,

Forum rules state you must show some work in order to get homework help. What have you tried so far? What equations/concepts do you think you might use? What are you stuck on?
 
  • #3
Well, for the 2nd question, I think we can use the equation: (Mice)(Lf) + (Mw)(Cw)(Change in Temp) = (Ms)(Lv) + (Msw)(Cw)(Change in Temp) since heat lost will be equal to heat gained. But I am not sure. Is this right?

For the 3rd question, can we use the equation: (Mass of water)(Specific heat capacity of water)(Temp of water) = (mass of steam)(Latent heat of vaporization)

Can anyone help
 

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