Odpowiedzi

2010-03-29T18:51:27+02:00
Dane:
m2 = 0,5 kg
t2 = 20*C
m1 = 1kg
t1 = 20 *C
m3 = ?
t3 = 100 * C

D-delta

Delta t1= 60*C-20*C= 40*C
Delta t2=60 * -20*=40*C
Delta t3=100*C-60*C=40*c
cw wody=4200j/kg*C
cw aluminium=900 J/kg*C

Q odebrane = Q pobrane
m1*cw1*DeltaT1+m2*Cw2*DeltaT2=m3*cw3 * DeltaT3
-m3*cw3*DT3=-m1*cw1*DT1-m2*cw2*DT2

-m3= (-m1*cw1*DT1 - m2*cw2*DT2) : Cw3*DT3
-m3= (-1kg*4200 J/kg*C * 40 C- 0,5kg * 900J/kg*C * 40 C) : 4200J/kg*C * 40 C
-m3=(-168000J-18000J):168000 J/kg
-m3=(-166000 J ) :168000 J/Kg
-m3=-1.1 kg
m3=1,1 kg