Oblicz ile ciepła trzeba dostarczyć lodowej brylce o masie 2 kg, aby otrzymać wodę o temp. 30 °C.
Temperatura lodu - 3°C.
Temperatura lodu - 3°C.
Isoko
![Dane:\\m=2[kg]\\T_p=-3[^oC]\\T_k=30[^oC]\\C_w_l_o_d_u = 2100 \frac{J}{kg*^oC }\\C_t = 334000 \frac{J}{kg}\\C_w_w_o_d_y = 4200 \frac{J}{kg*^oC }\\\\Szukane :\\Q = ? \\\\rozwiazanie :\\Q = Q_1+Q_2+Q_3\\\\Q_1 = C_w_l_o_d_u * m *\Delta T_1\\Q_2=C_t*m\\Q_3=C_w_w_o_d_y * m *\Delta T_2 \\\\ \Delta T_1 = 3[^oC]\\\Delta T_2 = 30 [^oC ] \\\\Q_1 = 2100 * 2 * 3 = 12 600 [J ]= 12,6[kJ] \\Q_2 = 334 000 * 2 = 668 000[ J] = 668 [kJ] \\Q_3=4200 * 30 * 2 = 252000[J] = 252 [kJ] \\\\Q = 12,6[kJ] + 668[kJ]+252[kJ]](https://tex.z-dn.net/?f=Dane%3A%5C%5Cm%3D2%5Bkg%5D%5C%5CT_p%3D-3%5B%5EoC%5D%5C%5CT_k%3D30%5B%5EoC%5D%5C%5CC_w_l_o_d_u+%3D+2100++%5Cfrac%7BJ%7D%7Bkg%2A%5EoC+%7D%5C%5CC_t+%3D+334000+%5Cfrac%7BJ%7D%7Bkg%7D%5C%5CC_w_w_o_d_y+%3D+4200+%5Cfrac%7BJ%7D%7Bkg%2A%5EoC+%7D%5C%5C%5C%5CSzukane+%3A%5C%5CQ+%3D+%3F+%5C%5C%5C%5Crozwiazanie+%3A%5C%5CQ+%3D+Q_1%2BQ_2%2BQ_3%5C%5C%5C%5CQ_1+%3D+C_w_l_o_d_u+%2A+m+%2A%5CDelta+T_1%5C%5CQ_2%3DC_t%2Am%5C%5CQ_3%3DC_w_w_o_d_y+%2A+m+%2A%5CDelta+T_2+%5C%5C%5C%5C+++%5CDelta+T_1+%3D+3%5B%5EoC%5D%5C%5C%5CDelta+T_2+%3D+30+%5B%5EoC+%5D+%5C%5C%5C%5CQ_1+%3D+2100+%2A+2+%2A+3+%3D+12+600+%5BJ+%5D%3D+12%2C6%5BkJ%5D+%5C%5CQ_2+%3D+334+000+%2A+2+%3D+668+000%5B+J%5D+%3D+668+%5BkJ%5D+%5C%5CQ_3%3D4200+%2A+30+%2A+2+%3D+252000%5BJ%5D+%3D+252+%5BkJ%5D+%5C%5C%5C%5CQ+%3D+12%2C6%5BkJ%5D+%2B+668%5BkJ%5D%2B252%5BkJ%5D+ "Dane:\\m=2[kg]\\T_p=-3[^oC]\\T_k=30[^oC]\\C_w_l_o_d_u = 2100 \frac{J}{kg*^oC }\\C_t = 334000 \frac{J}{kg}\\C_w_w_o_d_y = 4200 \frac{J}{kg*^oC }\\\\Szukane :\\Q = ? \\\\rozwiazanie :\\Q = Q_1+Q_2+Q_3\\\\Q_1 = C_w_l_o_d_u * m *\Delta T_1\\Q_2=C_t*m\\Q_3=C_w_w_o_d_y * m *\Delta T_2 \\\\ \Delta T_1 = 3[^oC]\\\Delta T_2 = 30 [^oC ] \\\\Q_1 = 2100 * 2 * 3 = 12 600 [J ]= 12,6[kJ] \\Q_2 = 334 000 * 2 = 668 000[ J] = 668 [kJ] \\Q_3=4200 * 30 * 2 = 252000[J] = 252 [kJ] \\\\Q = 12,6[kJ] + 668[kJ]+252[kJ]")
![\\Q = 932,6 [kJ]](https://tex.z-dn.net/?f=%5C%5CQ+%3D+932%2C6+%5BkJ%5D "\\Q = 932,6 [kJ]")
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