Odpowiedź :
Odpowiedź:
m1=2kg
t1=10°C
m2=1kg
t2=90°C
m1(tk-t1)=m2(t2-tk)
2(tk-10)=1(90-tk)
2tk-20=90-tk
2tk+tk=90+20
3tk=110
tk=36,6°
[tex]Dane:\\m_1 = 2 \ kg \ \ (1 \ l \ wody = 1 \ kg)\\t_1 = 10^{o}C\\m_2 = 1 \ kg\\t_2 = 90^{o}C\\Szukane:\\t_{k} = ?\\\\Rozwiazanie\\\\Q_{pobrane} = Q_{oddane}\\\\m_1c(t_{k}-t_1) = m_2c(t_2-t_{k}) \ \ /:c\\\\m_1(t_{k}-t_1) = m_2(t_2-t_{k})\\\\m_1t_{k}-m_1t_1 = m_2t_2-m_2t_{k}\\\\m_1t_{k} + m_2t_{k} = m_1t_1+m_2t_2\\\\t_{k}(m_1+m_2) = m_1t_1+m_2t_2 \ \ /:(m_1+m_2)\\\\t_{k} = \frac{m_1t_1+m_2t_2}{m_1+m_2}[/tex]
[tex]t_{k} = \frac{2 \ kg\cdot10^{o}C+1 \ kg\cdot90^{o}C}{2 \ kg + 1 \ kg}\\\\\boxed{t_{k} \approx 36,7^{o}C}[/tex]