Equilibrium temperature of a mixture

  • Thread starter frazdaz
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  • #1
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Homework Statement


2vnhu85.png


Homework Equations


[tex]Q = m c \Delta T \\
Q = m L_v \\
Q_{in} = Q_{out}[/tex]

The Attempt at a Solution


[tex]m_s c_s (100 - 8) + m_s L_v = m_s c_s (100 - T) + m_m c_s (T) \\
5778.06 + 33900 = 5778.06 - 62.805 T + 418.7 T \\
33900 = 355.895 T \\
∴ T = 95.25°C
[/tex]
Answer should be 90. This particular working is using 8°C as a base but I've with absolute temperatures and still got the same answer so I'm guessing my logic is flawed?
Thanks
 

Answers and Replies

  • #2
TSny
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Your first equation doesn't appear to be set up correctly. The left side has two terms and the right side has two terms.

Can you describe in words the meaning of each of the four terms in the equation?
 
  • #3
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Your first equation doesn't appear to be set up correctly. The left side has two terms and the right side has two terms.

Can you describe in words the meaning of each of the four terms in the equation?
Qin = extra thermal energy the steam has over the milk + heat to be released when the steam condenses
Qout = heat lost from the steam + heat gained by milk
 
  • #4
SteamKing
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The term

m[itex]_{s}[/itex]c[itex]_{s}[/itex](100-8)

implies that the mass of steam cools from 100 C to the temperature of the milk at 8 C after the steam condenses. This is clearly not the case, as the mixture will assume some as yet unknown temperature, which is higher than 8 C once thermal equilibrium is reached.

Rather than use Q[itex]_{in}[/itex] = Q[itex]_{out}[/itex], which is somewhat vague for this type of problem, why not use instead Q[itex]_{lost-from-steam}[/itex] = Q[itex]_{gained-by-milk}[/itex]
 
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