Final temperature of a mixture of water and ice in a copper containerLets say I have ice of mass m i and initial temperature T i and specific heat s i . And I have water of mass m w and initial temperature T w and specific heat s w . I have put both this water and ice in a copper container with mass m c u , specific heat s c u and initial temperature T w . What will be the final temperature T of the system ? ( The Latent heat of freezing of water is L f )Here , Q 1 = m i s i ( T − T i ) + m i L f Q 2 = m w s w ( T − T w ) + m w L f + m c u s c u ( T − T w )According to Calorimetry , Q 1 = Q 2 m i s i ( T − T i ) + m i L f = Q 2 = m w s w ( T − T w ) + m w L f + m c u s c u ( T − T w )I can use this equation to calculate T , But then I ran into a problem. In this case , I don't know which is happening , the freezing of water , or the melting of ice , so i can't just put both m w L f and m i L f in the equation and call it done. I might have to remove one of them depending on which thing is not happening (freezing or melting) .To make matters worse , The water or ice may just freeze or melt partially. Is there another equation that can solve for the final temperature while also taking all of this into account?

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Final temperature of a mixture of water and ice in a copper containerLets say I have ice of mass       m    i   and initial temperature       T    i   and specific heat       s    i   . And I have water of mass       m    w   and initial temperature       T    w   and specific heat       s    w   . I have put both this water and ice in a copper container with mass       m          c      u       , specific heat       s          c      u       and initial temperature       T    w   . What will be the final temperature   T of the system ? ( The Latent heat of freezing of water is       L    f   )Here ,      Q    1    =      m    i        s    i    (  T  −      T    i    )  +      m    i        L    f        Q    2    =      m    w        s    w    (  T  −      T    w    )  +      m    w        L    f    +      m          c      u            s          c      u        (  T  −      T    w    )According to Calorimetry ,      Q    1    =      Q    2        m    i        s    i    (  T  −      T    i    )  +      m    i        L    f    =      Q    2    =      m    w        s    w    (  T  −      T    w    )  +      m    w        L    f    +      m          c      u            s          c      u        (  T  −      T    w    )I can use this equation to calculate   T , But then I ran into a problem. In this case , I don&#039;t know which is happening , the freezing of water , or the melting of ice , so i can&#039;t just put both       m    w        L    f   and       m    i        L    f   in the equation and call it done. I might have to remove one of them depending on which thing is not happening (freezing or melting) .To make matters worse , The water or ice may just freeze or melt partially. Is there another equation that can solve for the final temperature while also taking all of this into account?

Adeline Shah · Accepted Answer

You actually have to consider all 3 cases: all the ice melts, all the water freezes, or you end up with a mixture of ice and water. Also, it is much more convenient to work with the first law of thermodynamics in solving this. Because we are assuming that this system is isolated, it tells us that the change in internal energy of the system is zero.Let       m          w      i       and       m          i      i       be the masses of water and ice initially and       m          w      f       and       m          i      f       be the masses of water and ice finally. Also, let the reference state for internal energy be water at 0 C. Then the specific internal energy of the water and ice initially are      u          w      i        =      s    w    (      T    w    −  0  )      u          i      i        =  −      L    f    −      s    i    (  0  −      T    i    )and the specific internal energy of water and ice finally are      u          w      f        =      s    w    (  T  −  0  )      u          i      f        =  −      L    f    −      s    i    (  0  −  T  )and the specific internal energy of water and ice finally are      u          w      f        =      s    w    (  T  −  0  )      u          i      f        =  −      L    f    −      s    i    (  0  −  T  )So setting the overall internal energy of the system initially to the overall internal energy of the system finally gives:      m          w      i            s    w        T          w        +      m          i      i        (  −      L    f    +      s    i        T    i    )  +      m          c      u            s          c      u            T    w    =      m          w      f            s    w    T  +      m          i      f        (  −      L    f    +      s    i    T  )  +      m          c      u            s          c      u        TThis equation is subject to the constraint that the total mass of water and ice is constant:      m          w      i        +      m          i      i        =      m          w      f        +      m          i      f      The three possible cases are then as follows:1.If water and ice are both present finally, then T = 02.If only water is present finally, then       m          i      f        =  03.If only ice is present finally, then       m          w      f        =  0

Jayden Mckay · Accepted Answer

Assume the end-temperature to be   T. Your expression for       Q    1   is then wrong.The ice will heat from       T    i   to   0, then absorb latent heat of melting, then the melted water will heat to   T, so we have:      Q    1    =      m    1        s    i    (      T    i    −  0  )  +      m    1        L    f    +      m    w        s    w    (  0  −  T  )

Final temperature of a mixture of water and ice in a copper container Lets say I have ice of mass

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