Odpowiedź :
Odpowiedź:
m1=2kg
t1=20°C
t2=100°C
t3=40°C
Q1=m1*(t3-t1)
Q2=m2*(t2-t3)
Q=Q2
m1*(t3-t1)=m2*(t2-t3)
m2=m1*(t3-t1)/(t2-t3)
m2=2kg*(40-20)°C/(100-40)°C
m2≈0,67kg
[tex]Dane:\\m = 2 \ kg\\T_1 = 20^{o}C\\T_2 = 100^{o}C\\T_{k} = 40^{o}C\\Szukane:\\m_2 = ?\\\\Rozwiazanie\\\\Q_{pobrane} = Q_{oddane}\\\\m_1c_1(T_{k}-T_1) = m_2c_2(T_2-T_{k})\\\\c_1 = c_2 = c\\\\m_1c(T_{k}-T_1) = m_2c(T_2-T_{k}) \ \ /:c(T_2-T_{k})\\\\m_2 = {m_1\cdot\frac{(T_{k}-T_1)}{T_2-T_{k}}}\\\\m_2 = 2 \ kg\cdot\frac{40^{o}C - 20^{o}C}{100^{^o}C - 40^{o}C}}\\\\m_2 = 2 \ kg\cdot\frac{20^{o}C}{60^{o}C}\\\\\boxed{m_2=0,(6) \ kg \approx0,67 \ kg}[/tex]
Odp. Należy dolać ok. 0,67 kg wrzątku.