Determine the final temperature of the mixture

  • Thread starter Stomper123
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  • #1

Homework Statement


A student places 21.4 grams of ice at 0.0C and 13.1 grams of steam at 100.0C in a sealed and insulated container. Determine the final temperature of the mixture.
m(ice)= 21.4 g or 0.0214 kg
m(steam)= 13.1 g or 0.0131 kg
Ti(ice)= 278.15 K
Ti(steam)= 378.15 K
Tf(ice)=Tf(steam)
Latent heat of fusion: 334 kJ/Kg
Latent heat of vaporization 2257 kJ/kg

Homework Equations


Q=mcΔT
Q=mL
Qt(hot)= -Qt(cold)

The Attempt at a Solution


I did it on paper and attempted to solve for T2 using latent heat of fusion on one side and the vaporisation on the other side but ended up getting 57.74K which I know is impossible. I think one of the objects doesn't fully change state but I don't know how to do those kinds of problems.
 

Answers and Replies

  • #2
cepheid
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I think what you can do is look at the difference between the energy required to melt that amount of ice, and the energy required to condense that amount of steam. The former is less than the latter. Once enough energy has been transferred to melt all of the ice, the steam is still condensing, and the balance/remainder is used to heat that 0.0 C liquid water. Once the condensing is finished, you'll have some newly condensed water at just under 100 C, and some other liquid water at some temperature that depends on how much heating took place after melting (which depends on this difference/balance/remainder of energy I referred to in bold above). Now you can just use Q = mc delta T to find the final equilibrium temperature from this starting point of two amounts of liquid water at two different temperatures.
 
  • #3
SteamKing
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Check your absolute temperature values. I think you will find them to be slightly off.
 

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