Somebody would like to evaporate 2 kg alcohol. if alcohol first temperature is -100 celcius, decide the amount of heat needed ?
Data: alcohol melting point : -97 celcius
alcohol boiling point: 78 celcius
alcohol melting heat: 69.000 J/kg
alcohol specific heat: 2400 J/Kg celcius
alcohol vapor heat: 1.100.000 J/Kg celcius

1
itu soalnya ga salah ya?
vapor heat (kalor uap) satuannya J/kg
ga ad celciusnya
coba langsung jak pake cara kau
ga sopan sekali --a

Jawabanmu

2014-03-05T16:24:50+07:00
Dari -100 ke -97
ΔT=3⁰C

Q1 = 2*3*2400
=14400 J
=14,4 KJ
mencair
Q2 = 2*69000
= 138000 J
=138KJ
dari -97 ke 78
ΔT = 175⁰C
Q3 = 2*175*2400
= 840000J
=840KJ
menguap
Q4 = 2* 1100000
=2200000J
=2200KJ

total panas
Q1+Q2+Q3+Q4
=14,4+138+840+2200
=3192,4KJ