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The question simply says “calculate the temperature of hot air in kelvin” on a data calculation section for this lab. We recorded:

  • volume change on cooling = 170 mL
  • original volume of hot air = 20 mL
  • volume of the cooled air = 150 mL
  • temperature of heated water = 100 °C
  • temperature of cool water = 20.4 °C

and that is all.

The lab was focused on Charles’ law, namely to predict how much air should contract when cooled from 100 °C to room temperature. We took a flask and inserted a rubber stopper with glass tubing through the center. We then heated the flask in a water bath until the water boiled, then keeping a finger over the top of the glass tubing we inverted the flask in a “cool” water bath and held it there for 10 minutes. Then we equalized the pressure within the flask with the atmospheric pressure by raising the flask until the water level inside equaled the outside.

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  • $\begingroup$ Should the volume change not be 20 mL and the hot volume 170 mL. Otherwise it doesn't make any sense? $\endgroup$ – bon Jun 14 '15 at 19:12
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You already know the temperature of the hot air in Celsius: 100 degrees

Using the equation Kelvin = Celsius + (273.15 degrees), you would get 100 degrees + 273.15 degrees, or 373.15 degrees.

I am assuming that this is a preliminary step that is required in order to find the temperature to plug into Charles' law (which must be in Kelvin), in which case you can use the formula I mentioned above for your other calculations as well.

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