Calculate the heat content in a two tonne (2,000 kg) iron magnet at 300 K. How...
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Calculate the heat content in a two tonne (2,000 kg) iron magnet at 300 K. How much liquid nitrogen is required to cool this magnet to 80 K? How much liquid helium is required to cool this magnet from 80 to 4 K? (Hint: Assume that the internal energy change is entirely absorbed by the liquid resulting is a mass of vapor. Use the Debye model to calculate the change in internal energy, (Fe) = 460 K; hfg(He@ 4 K)=21 kJ/kg; hfg (N₂ @ 80 K=200 kJ/kg)). Calculate the heat content in a two tonne (2,000 kg) iron magnet at 300 K. How much liquid nitrogen is required to cool this magnet to 80 K? How much liquid helium is required to cool this magnet from 80 to 4 K? (Hint: Assume that the internal energy change is entirely absorbed by the liquid resulting is a mass of vapor. Use the Debye model to calculate the change in internal energy, (Fe) = 460 K; hfg(He@ 4 K)=21 kJ/kg; hfg (N₂ @ 80 K=200 kJ/kg)).
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Mass of iron magnet 2000 kg Initial temperature 300 K Final temperature for liquid nitrogen cooling ... View the full answer
Related Book For
Fundamentals of Thermodynamics
ISBN: 978-0471152323
6th edition
Authors: Richard E. Sonntag, Claus Borgnakke, Gordon J. Van Wylen
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