Ile ciepła należy dostarczyć, aby podgrzać 1 kg wody o w miedzianym garnku o masie 0,65 kg od 20°C do temp. wrzenia?
Cw wody - ok. 4200
Cw miedzi - 385
ma wyjść Q=355 kJ
Cw wody - ok. 4200
Cw miedzi - 385
ma wyjść Q=355 kJ
Isoko
![Dane:\\m_1=1[kg]\\m_2=0,65[kg]\\T_p=20[^oC]\\T_k=100[^oC]\\Cw_1=4200[ \frac{J}{kg\cdot^oC}]\\Cw_2=385[ \frac{J}{kg\cdot^oC} ] \\\\Szukane :\\Q = ? \\\\Rozwiazanie :\\Q = Q_1 + Q_2 \\\\Q = \Delta T ( Cw_1\cdot m_1 + Cw_2 \cdot m_2)\toQ=(T_k-T_p)( Cw_1\cdot m_1 + Cw_2 \cdot m_2) \\\\Q = (100-20)(1*4200+0,65*385) = 80\cdot 2250,25 \\Q \approx 356000 [J]=356[kJ]](https://tex.z-dn.net/?f=Dane%3A%5C%5Cm_1%3D1%5Bkg%5D%5C%5Cm_2%3D0%2C65%5Bkg%5D%5C%5CT_p%3D20%5B%5EoC%5D%5C%5CT_k%3D100%5B%5EoC%5D%5C%5CCw_1%3D4200%5B+%5Cfrac%7BJ%7D%7Bkg%5Ccdot%5EoC%7D%5D%5C%5CCw_2%3D385%5B+%5Cfrac%7BJ%7D%7Bkg%5Ccdot%5EoC%7D+%5D+%5C%5C%5C%5CSzukane+%3A%5C%5CQ+%3D+%3F+%5C%5C%5C%5CRozwiazanie+%3A%5C%5CQ+%3D+Q_1+%2B+Q_2+%5C%5C%5C%5CQ+%3D+%5CDelta+T+%28+Cw_1%5Ccdot+m_1+%2B+Cw_2+%5Ccdot+m_2%29%5CtoQ%3D%28T_k-T_p%29%28+Cw_1%5Ccdot+m_1+%2B+Cw_2+%5Ccdot+m_2%29+%5C%5C%5C%5CQ+%3D++%28100-20%29%281%2A4200%2B0%2C65%2A385%29+%3D+80%5Ccdot+2250%2C25+%5C%5CQ+%5Capprox+356000+%5BJ%5D%3D356%5BkJ%5D "Dane:\\m_1=1[kg]\\m_2=0,65[kg]\\T_p=20[^oC]\\T_k=100[^oC]\\Cw_1=4200[ \frac{J}{kg\cdot^oC}]\\Cw_2=385[ \frac{J}{kg\cdot^oC} ] \\\\Szukane :\\Q = ? \\\\Rozwiazanie :\\Q = Q_1 + Q_2 \\\\Q = \Delta T ( Cw_1\cdot m_1 + Cw_2 \cdot m_2)\toQ=(T_k-T_p)( Cw_1\cdot m_1 + Cw_2 \cdot m_2) \\\\Q = (100-20)(1*4200+0,65*385) = 80\cdot 2250,25 \\Q \approx 356000 [J]=356[kJ]")
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