Do 0 5 , kg wody o temperaturze 20°C dolano 1 3 litra wody o temperaturze 60°C. Oblicz temperaturê koñcow¹ wody.

Dane: m₁=0,5kg t₁=20^C m₂=⅓kg t₂=60^C Szukane: t₃=? Wzór: Q₁=Q₂ m₁*c*(t₃-t₁)=m₂*c*(t₂-t₃)/:c m₁*(t₃-t₁)=m₂*(t₂-t₃) m₁t₃-m₁t₁=m₂t₂-m₂t₃ m₁t₃+m₂t₃=m₂t₂+m₁t₁ t₃*(m₁+m₂)=m₂t₂+m₁t₁/:(m₁+m₂) t₃=m₂t₂+m₁t₁:(m₁+m₂) Rozwiązanie: t₃=⅓kg*60^C+0,5kg*20^C:(0,5kg+⅓kg) t₃=20kg*^C+10kg*^C:(5/6kg) t₃=30kg*^C:5/6kg t₃=36^C Odp. Temp. końcowa wyniesie 36^C.   Pozdrawiam. ;)

dane: m1 = 0,5 kg t1 = 20* C m2 = 1/3 kg (1 l = 1 kg) t2 = 60* C C = 4200 J/kg*C szukane: tk = ?   Q pobrane = Q oddane m1 *C*(tk-t1) = m2 *C*(t2-tk)  ?:C m1(tk-t1) = m2(t2-tk) m1*tk - m1*t1 = m2*t2 - m2*tk m1*tk + m2*tk = m2*t2 + m1*t1 tk(m1+m2) = m2*t2 + m1*t1   /:(m1+m2) tk = (m2t2+m1t1)/(m1+m2) tk = (1/3kg*60*C + 0,5kg*20*C)/(1/2+1/3)kg tk = 36* C =========