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1 kg of ice at -20°C is mixed with 1kg steam at 200°C.....

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  1. Mar 8, 2016 #1
    • Moved from a non-homework forum section, so missing the HW template
    1 kg of ice at -20°C is mixed with 1kg steam at 200°C. Then find equilibrium temperature and mixture content.

    Please DON'T use Joules as unit while answering the question.

    Converting 1 kg steam at 200°C to 1 kg water at 0°C requires : (1* 0.5*100) + ( 1*540) + (1* 1 * 100 ) k cal heat = 690 kcal heat.

    Converting 1 kg ice at -20°C to 1 kg water at 0°C requires : (0.5 * 1* 20 ) + (1* 80) kcal heat = 90 kcal heat.

    Heat released by converting 1kg steam to 1 kg water at °C will be used to provide ice the required heat.

    Therefore, heat remaining = 690-90 kcal=600 kcal heat

    Now we have 600 kcal heat and 2kg of water at 0°C.

    ⇒600= 2* 1 * change in temp ⇒ change in temp. = 300 °C

    But that is not the answer. The solution in my textbook says that the heat required to ice is less than the heat supplied by steam. But I don't see that happening.
    In the method I have used, we convert the substance at relatively higher temperature to the same substance at relatively lower temperature as I have converted steam to water at 0°C here, and then use Q= m*s* ΔT to find the change in temperature. I don't know why but it does not seem to work here and the answer does not seem right to me either. What is wrong in this procedure? Please don't just give solution to the question ; tell me the flaw in my procedure.
     
    Last edited: Mar 8, 2016
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  3. Mar 8, 2016 #2

    BvU

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    Please post in the homework forum :smile:

    What happens when you heat up water ? Does the temperature keep going up ?

     
  4. Mar 8, 2016 #3
    I have used the values of latent heat of fusion and vaporisation in my calculations.
     
  5. Mar 8, 2016 #4

    BvU

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    Please answer the question
     
  6. Mar 8, 2016 #5
    When we heat the water the temperature will keep on rising till 100 °C and will stop increasing after that until all the water at 100°C has been converted to steam at 100°C(latent heat of vaporisation). Then the temperature starts rising again.
     
  7. Mar 8, 2016 #6

    BvU

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    Very good. So from the 600 kcal you have "available", 200 can be used to bring the kettle to the boil. The remainder causes water to evaporate. How much ? Your call !
     
  8. Mar 8, 2016 #7
    I don't understand why are we using those 200 kcal to boil water. Haven't we already taken into account all that heat while arriving at 600 kcal ?
     
  9. Mar 8, 2016 #8

    BvU

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    Yes we have. It's left over. Not what you would call a decent equilibrium ....

    You calculated how much heat would be left over if the whole lot was brought to liquid water at 0 C. Apparently that isn't the equilibrium temperature !
     
  10. Mar 8, 2016 #9
    Okay, so 2kg water will keep on increasing temperature until it reaches 100°C. Heat required = 2* 1* 100 kcal = 200 kcal
    Now we have 2 kg of water at 100°C but only 400 kcal of heat to convert it into steam. So only some of the water will convert into steam.
    So only, (400/540)=0.74 gm water will convert to steam.
    Total steam = 0.74 kg
    Total water = 2-0.74 kg =1.25 kg
    This is the correct answer as well.
    I can't thank you enough.
     
  11. Mar 8, 2016 #10

    BvU

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    You're welcome.
     
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