Suppose that 18g ofsteam at 100 deg C interacts with 50 g of ice at 0 deg C. Wha

Question

Suppose that 18g ofsteam at 100 deg C interacts with 50 g of ice at 0 deg C.  What is the final temerature? Explain your reasoning. (HInt: Consider whether al the steam must condense in order to melt the ice.)

Attempt:

I tried 0=(heat for steam to equallibrium)-(heat of fusion)-(heat of water equallibrium)

          0 = (18g)(1.996 J/g/K)(T_f - 100) - (50g)(333 J/g) - (50g)(4.187 J/g/k)(T_f - 0)

          T_f=-116.72   <--- not within 0 to 100 range

So my professor says to use the following.

         where m equals that mass of steam converted to water

m(L_(vap)) = (heat of fusion) + (energy to heat 50g of water 100 degrees)

m(2256) = 50g(333 J/g) + 50g(4.187 J/g/K)(100 deg C - 0 deg C)

        m=16.66g

        mass of steam left over equals 1.34g  

So I have 3 questions.

1) I thought that there is a latent heat for vaporization not condensation.  Is it the same as vaporization?

2) How do we know that all the water is heat to 100 deg C?  

3)Setting the second formula up, we are assuming that not all of the vapor condensed.  

    How do we come to this conclusion?

Explanation / Answer

Heat of condensation = Heat of vaporization Heat of fusion = Heat of melting These are "reversible" thermodynamic processes. Heat lost by condensing 18 gm steam = 18 gm * 540 cal/gm = 9720 cal Heat required to melt 50 g ice = 50 g * 80 cal/g = 4000 cal Heat required to raise 50 g water to 100 deg = 50 * 100 = 5000 Total heat required to melt ice and heat to 100 deg = 9000 cal So the final temperature is 100 deg 9000 / 540 = 16.7 g is the amount of steam needed to condense to reach the final equilibrium temperature of 100 deg.

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