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Metal in water thermal equilibrium? Help

  1. Nov 3, 2006 #1
    Metal in water thermal equilibrium?? Help

    Determining temperature of an oven.

    Guy puts a copper bar with a mass of 5.0 kg in the oven and puts an identical bar in a well-insulated 20.0 liter vessel containing 5.00 L of liquid water and the rest saturated steam at 760 mmHg. Waits until bars reach thermal equilibrium with their surroundings, then quickly takes the first bar out of the oven, removes the second bar from the vessel, drops the first bar in its place, covers the vessel tightly, waits for the system to come to equilibrium, and records the pressure gauge reading inside the vessel. It is 50.1 mmHg. Specific gravity of copper is 8.92. Specific internal energy of copper is given by U hat (kJ/kg) = 0.36*T(degrees celcius). to calculate oven temperature.

    a) assume bar can transfer from oven to vessel without losing heat. What temperature is the oven? How many grams H2O evaporate?
    b) bar actually lost 8.3 kJ of heat b/w oven and the vessel. Whatis the true oven temperature.
  2. jcsd
  3. Nov 4, 2006 #2
    Anyone have an idea? We worked on this shet for 3 hours, but couldn't relate the information and pressure stuff. Involves energy equil. and stuff.
  4. Nov 4, 2006 #3


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    The fact that the bars are identical removes the volume from consideration of the change in pressure in the vessels, so any change in pressure is due to thermal effects (heat transfer) from the block.

    There is some energy content in the heated bar, which is related to the mass, specific heat and temperature of the bar.

    Then there is the initial conditions in the vessel, water and sat steam at 760 mm Hg, which is 1 atm (I expect absolute pressure by the wording) so the temperature of the water must be 100°C. Now the gage reading (usually gage pressure) is 50.1 mm Hg, which is equiavlent to 760+50.1 mmHg absolute. So there is a differential pressure of 50.1 mmHg, which comes from the heat.

    The heat is transfered from the block to the water - and some of the 5.0 l of water is converted to steam - hence the rise in pressure. There is heat of vaporization involved. There is change of phase in a constant volume (20 l) process.

    We know that that the heated block must have a temperature greater than

    Heat content (thermal energy) = m cp(T) or
    Change in heat content = m cp([itex]\Delta[/itex]T)

    See also - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html
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