No, I didn't read past the first line. Steam is not liquid water; therefore, you may not use the specific heat of water. You must use the specific heat of steam. THEN, it doesn't go from 125 to zero. It goes from 125 to 100.
In the initial post I told you what we were converting; i.e., from steam at 125 to steam at 100, then condense the vapor steam to liquid water at 100, then move form liquid water at 100 to liquid water at zero, then heat of fusion to freeze water to ice at zero, then move from ice at zero to ice at -125.
Hi, I just need to know if I finally got this exercise right.
Calculate the amount of heat in kilojoules released when 50 grams of steam at 125 Celsius are converted to 50 grams of ice at -125 Celsius.
so,
qw= 50*4.184J/g C (125 C-o)= 26.15 x 10^3
qi= 50*2.09*(-125)=-13.065 KJ
q= 50*1.860 (-125)= -11.63 KJ
q1+q2=q3= -11.92 kJ
please let me know if this way is correct thanks
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