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U-Tube and Heat

  • #1
I really suck at these problems. I dont know what I'm doing wrong. Please help.

1. A U-tube contains liquid of unknown density. An oil of density 785 kg/m3 is poured into one arm of the tube until the oil column is 15 cm high. The oil-air interface is then 7 cm above the liquid level in the other arm of the U-tube. Find the density of the liquid.

I know that P1=P2
I said P1 = Patm + (density of oil)(g)(.15m) + (density of unknown)(g)(.07m)
P2 + Patm + (density of unknown)(g)(.15)
When I solved for the density, I got and answer of 1471.875.
Is this right? I get the feeling I did something wrong. Please help!

2. A well-insulated styrofoam bucket contains 143 g of ice at 0oC. If 21 g of steam at 100°C is injected into the bucket, what is the final equilibrium temperature of the system?(cwater=4186 J/kg.oC, Lf=33.5x104 J/kg, Lv=22.6x105 J/kg)

Qlost= steam
Qgain = ice

qlost=q gain

(1) M ice * C ice * (Teq-0) + M ice * Lf + M ice * C water * (Teq - 0)
(2) M steam * C steam * (Teq - 100) + M steam * Lv + M steam C water ( 100 - 0)

I equated the two equations, and then solved for Teq. I messed up somewhere, I think with temeperatures. Any help will be appreciated.
 

Answers and Replies

  • #2
can anyone please help me =(
 
  • #3
alphysicist
Homework Helper
2,238
1
Hi peaches1221,

#1-Your approach here gives the right answer. However, I think it might be better to think about it this way:

On the side with the oil, the pressure at the top of the oil is atmospheric pressure. At the bottom of the oil the pressure is

[tex]
P_1 = P_{\rm atm} + \rho_{\rm oil}g(0.15)
[/tex]

Now find the pressure of the unknown liquid at the same vertical depth as the bottom of the oil. The pressure at the top of the liquid is atmospheric pressure. The top of the liquid is 7cm below the top of the oil, so you have to go down 8cm into the liquid to reach the same vertical depth. So

[tex]
P_2 = P_{\rm atm} + \rho_{\rm liq} g(0.08)
[/tex]

Then set [itex]P_1=P_2[/itex] to find [itex]\rho_{\rm liq}[/itex]. (But like I said, this is equivalent to what you did.)


#2- For the ice, since it starts out at its melting point temperature, there are only two processes: it melts, and then it heats up to the equilibrium temperature. So there should only be two terms in the q gain equation.

Similarly for the steam; since it starts out at 100 degrees C, it will only have two processes: it turns to water, and then cools down to the equilibrium temperature.
 

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