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Making iced tea

  1. Dec 8, 2012 #1
    a person makes a quantity of iced tea by mixing 500 g of hot tea (essentially water) with an equal mass of ice at its melting point. if the initial hot tea is at room temperature of (a) 90 degrees celsius and (b) 70 degrees celsius, what are the temperature and mass of the remaining ice when the tea and ice reach a common temperature? Neglect energy transfers with the environment.




    My attempt

    Latent heat of melting - 334 kJ/kg=L
    s=Specific heat water - 4.187 kJ/kgK
    from calorimetry principle,
    heat rejected by tea=heat used to melt ice
    a)Final temp=t degree
    all ice melts
    heat rejected=ms(90-t)=.5*4.187*(90-t)
    heat tken in ice=mL+mst=.5*334+.5*4.187*t
    so .5*4.187*(90-t)=.5*334+.5*4.187*t
    temp=5.11

    b)let m ice melts
    final temp=0
    heat rejected=ms(70-0)=.5*4.187*(70-0)=146.545
    heat taken in ice=mL=m*334
    so 146.545=m*334
    so m=.439 g
    so ice remains=.5-.439=.061


    ans :
    a)final temp=5.11 degree c ; no ice remains
    b)Final temp=0, .061 kg or 61.2 gram ice remain



    Can someone please tell me why in Part A 90 is being used as the final temperature and t is used as the initial temperature? Thank you! I am really confused about that part
     
  2. jcsd
  3. Dec 8, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Heat rejected is the negative of heat gained. To take care of the negative, the initial and final temperatures are switched in calculating the change in temperature for the heat rejected. That way, you get a positive number for the heat rejected.
     
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