Science

A peice of ice weighting 5g at -20 centigrade is put into 10g of ice at 30C. Assuming no heat is lost to the outside.Calculate the final tempature of the mixture?

asked by Om
  1. Ice at 30 deg C??? Not likely but anyway

    specific heat of ice (about 2 J/gram deg c) does not matter in the end here
    call it sh
    heat into cold ice = heat out of hot ice
    5 (T - -20) * sh = 10 (30-T) * sh
    5 T + 100 = 300 - 10 T
    15 T = 300
    T = 20
    BUT I doubt the whole thing
    suspect that 10 grams was water at 30 deg
    In that case
    Heat in to heat the ice up to zero
    5 (20)*sh
    Heat in to melt the ice
    5 * hm where hm is the heat of fusion of water
    Heat in to warm melted ice up to T
    5(T-0) * shw
    where shw is specific heat of water

    Heat out of 30 deg water
    10(30-T)*shw
    so
    5 (20)*sh + 5 * hm + 5(T-0) * shw = 10(30-T)*shw
    If T comes out below zero, then T = 0 and some of the 30 deg water froze

    posted by Damon

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