1 kg oleju o temp. 20 st dodano do 3 kg oleju o temp. 100 st. Jaka jest temp. końcowa mieszaniny

dane: m₁ = 1 kg t₁ = 20°C m₂ = 3 kg t₂ = 100°C szukane: tk = ?   Q pobrane = Q oddane m₁C(tk-t₁) = m₂C(t₂-tk)  /:C m₁(tk-t₁) = m₂(t₂-tk) m₁tk - m₁t₁ = m₂t₂ - m₂tk m₁tk + m₂tk = m₁t₁ + m₂t₂ tk(m₁+m₂) = m₁t₁+m₂t₂   /:(m₁+m₂) tk = (m₁t₁+m₂t₂)/(m₁+m₂) tk = (1kg*20°C + 3kg*100°C)/(1kg+3kg) = 320/4 °C tk = 80°C -------------