[tex]Dane:\\m_1 = 200 \ kg\\t_1 = 4^{o}C\\t_2 = 100^{o}C\\t_{k} = 21^{o}C\\Szukane:\\m_2 = ?[/tex]
Rozwiązanie
[tex]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}) \ \ \ |:(t_2-t_{k})\\\\m_2 = \frac{m_1(t_{k}-t_1)}{t_2-t_{k})}\\\\m_2 = \frac{200 \ kg\cdot(21^{o}C - 4^{o}C)}{100^{o}C - 21^{o}C}=\frac{200 \ kg\cdot17^{o}C}{79^{o}C}\\\\\boxed{m_2 \approx43 \ kg}[/tex]
Odp. Do brodzika trzeba jeszcze wlać ok. 43 kg wody o temperaturze 100°C.
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[tex]Dane:\\m_1 = 200 \ kg\\t_1 = 4^{o}C\\t_2 = 100^{o}C\\t_{k} = 21^{o}C\\Szukane:\\m_2 = ?[/tex]
Rozwiązanie
[tex]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}) \ \ \ |:(t_2-t_{k})\\\\m_2 = \frac{m_1(t_{k}-t_1)}{t_2-t_{k})}\\\\m_2 = \frac{200 \ kg\cdot(21^{o}C - 4^{o}C)}{100^{o}C - 21^{o}C}=\frac{200 \ kg\cdot17^{o}C}{79^{o}C}\\\\\boxed{m_2 \approx43 \ kg}[/tex]
Odp. Do brodzika trzeba jeszcze wlać ok. 43 kg wody o temperaturze 100°C.