jej moje ulubione zadania
1.
Dane:
m2=0,5 kg
t2=20C
m1=1kg
t1=20C
m3=?
t3=100 * C
D-delta ( bede sie posługiwala tym skrotem )
Delta T1 = 60C-20C = 40C
Delta T2=60 * - 20=40C
Delta T3=100C-60C=40c
cw wody = 4200j/kgC
cw aluminium = 900 J/kgC
Q odebrane = Q pobrane
m1cw1DeltaT1 + m2Cw2DeltaT2 = m3cw3 * DeltaT3
-m3cw3DT3 = -m1cw1DT1-m2cw2*DT2
-m3= (-m1cw1DT1 - m2cw2DT2) : Cw3DT3
-m3= (-1kg4200 J/kgC * 40 C - 0,5kg * 900J/kgC * 40 C) : 4200J/kg*C * 40 C
-m3=(-168000 J - 18000 J ) : 168000 J/kg
-m3=(-166000 J ) : 168000 J/Kg
-m3=-1.1 kg
m3=1,1 kg
Zaraz zrobie drugie :]