erobz
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A few notes, you don't need to convert by adding ##273##, just note that the temperatures used are to be absolute.Hak said:Here, I get totally stuck. Where do I go wrong?
What is being suggested (as an alternative approach) is at equilibrium ##\rho_{avg} = \rho_{out} ##. You can write that immediately as:
$$\frac{\rho_{in} V + m}{V} = \rho_{out} $$
Work from there by subbing the relationships you found earlier for ##\rho_{in}## as a function of ##h## and the relationship you are given for ##\rho_{out}##.
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