Dane: T₁=20°C m₁=0,5kg T₂=40°C m₂=ϱ(w)*v=1000kg/m³*⅓dm³=1000*1/3000dm³=⅓kg Szukane: T(k)=? Obl: Z bilansu cieplnego: Q₁=Q₂ c=Q/m*ΔT | *(m*ΔT) Q=c*m*ΔT c*m₁*(T(k)-T₁)=c*m₂*(T₂-T(k) | /c m₁*(T(k)-T₁)=m₂*(T₂-T(k) m₁T(k)-m₁T₁=m₂T₂-m₂T(k) | + m₂T(k) + m₁T₁ m₁T(k)+m₂T(k)=m₂T₂+m₁T₁ T(k)(m₁+m₂)=m₂T₂+m₁T₁ | /(m₁+m₂) T(k)=m₂T₂+m₁T₁/m₁+m₂ T(k)=⅓kg*40°C+0,5kg*20°C/0,5kg+⅓kg=13⅓+10/⅚[°C]=23⅓*6/5[°C]=28°C Odp: Temperatura końcowa wody wynosi 28°C.
m₁=0,5kg T₁=20°C m = 1/3 l = 1/3 kg T₂=40°C T(k)=? Korzystamy ze wzoru na bilans cieplny Q = m * c * T Q₁= Q₂ m1*c*T1 + m2*c*T2 = (m1+m2)*c*Tk // c - skracamy c (m1*T1 + m2*T2) / ( m1+m2) = Tk Tk = 0,5 kg *20C + 1/3 kg * 40C / 0,5 kg + 1/3 kg Tk = 10 kg*C + 40/3 kg*C / 5/6 kg Tk = (60/6 kg*C + 80/6 kg*C ) / 5/6 kg Tk = 140/6 kg*C / 5/6 kg = 28 C temp wody wyniesie 28 C T(k)=⅓kg*40°C+0,5kg*20°C/0,5kg+⅓kg=13⅓+10/⅚[°C]=23⅓*6/5[°C]=28°C Odp: Temperatura końcowa wody wynosi 28°C.
