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Dane:
m1=V1*d;
m2=V2*d.
T1=100[stop.];
T2=20[stop.].
m1=3[dm3]*1000[g/dm3]=3000[g]=3[kg].
m2=2[kg].
Z bilansu cieplnego otrzymujemy:
Tk=(m1T1+m2T2)/(m1+m2);
Tk=(3*100+2*20)/(3+2);
Tk=340/5=68[stop.].
Odp.Temperatura końcowa wody Tk=68[stop.].
1Lw = 1 kg
m1 =3 kg
Tp1 = 100 C
m2 =2 kg
Tp2 = 20 C
m1 * Cw * (Tk -Tp1) = m2 * Cw *(Tp2 -Tk)/:Cw
Z równania wyliczasz Tk:
m1(Tk - Tp1) = m2(Tp2 -Tk)
m1 * Tk - m1 * Tp1 = m2 * Tp2 -m2 * Tk / + m2 * Tk
m1 * Tk + m2 * Tk - m1 *Tp1 = m2 *Tp2 /+ m1 * Tp1
m1 *Tk + m2 * Tk - m1 *Tp1 + m1 * Tp1 = m2 *Tp2 + m1 * Tp1
Tk (m1 + m2) =m2 * Tp2 + m1 * Tp1/:(m1 +m2)
Tk =(m1 *Tp1 + m2 * Tp2):(m1 + m2)
Tk = ( 3 kg * 100 C + 2 kg * 20 C):(3 kg + 2 kg)
Tk =340/5 [C]
Tk = 68 stopni C
Odp. Temperatura końcowa wody równa jet 68 stopni C