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1. (2024·临夏州)某品牌电热水壶铭牌如表所示,该电热水壶内部简化电路图如图所示。该电热水壶正常工作时,将1.5L初温为$20^{\circ }C$的水加热至沸腾,用时6min,已知外界大气压为1标准大气压。
(1)只闭合$S_{1}$时,该电热水壶处于____(选填“保温”或“加热”)状态；
(2)求电阻$R_{2}$的阻值;
(3)求在上述过程中电热水壶烧水的效率。
额定电压:220V
额定频率:50Hz
加热功率:2000W
额定容量:1.7L
保温功率:40W
(1)只闭合$S_{1}$时,该电热水壶处于____(选填“保温”或“加热”)状态；
(2)求电阻$R_{2}$的阻值;
(3)求在上述过程中电热水壶烧水的效率。
额定电压:220V
额定频率:50Hz
加热功率:2000W
额定容量:1.7L
保温功率:40W
答案
(1)保温
(2)当开关 $ S_{1} $、$ S_{2} $ 闭合时，$ R_{1} = \frac{U_{额}^{2}}{P_{加}} = \frac{(220V)^{2}}{2000W} = 24.2\Omega $，
开关 $ S_{1} $ 闭合、$ S_{2} $ 断开时，$ R = \frac{U_{额}^{2}}{P_{保}} = \frac{(220V)^{2}}{40W} = 1210\Omega $，
$ R_{2} = R - R_{1} = 1210\Omega - 24.2\Omega = 1185.8\Omega $；
(3)$ m = \rho V = 1.0 \times 10^{3}kg/m^{3} \times 1.5 \times 10^{-3}m^{3} = 1.5kg $，
$ Q_{吸} = cm\Delta t = 4.2 \times 10^{3}J/(kg \cdot ^{\circ}C) \times 1.5kg \times (100^{\circ}C - 20^{\circ}C) = 5.04 \times 10^{5}J $，
$ W = P_{加热}t = 2000W \times 6 \times 60s = 7.2 \times 10^{5}J $，
$ \eta = \frac{Q_{吸}}{W} = \frac{5.04 \times 10^{5}J}{7.2 \times 10^{5}J} = 70\% $。
(2)当开关 $ S_{1} $、$ S_{2} $ 闭合时，$ R_{1} = \frac{U_{额}^{2}}{P_{加}} = \frac{(220V)^{2}}{2000W} = 24.2\Omega $，
开关 $ S_{1} $ 闭合、$ S_{2} $ 断开时，$ R = \frac{U_{额}^{2}}{P_{保}} = \frac{(220V)^{2}}{40W} = 1210\Omega $，
$ R_{2} = R - R_{1} = 1210\Omega - 24.2\Omega = 1185.8\Omega $；
(3)$ m = \rho V = 1.0 \times 10^{3}kg/m^{3} \times 1.5 \times 10^{-3}m^{3} = 1.5kg $，
$ Q_{吸} = cm\Delta t = 4.2 \times 10^{3}J/(kg \cdot ^{\circ}C) \times 1.5kg \times (100^{\circ}C - 20^{\circ}C) = 5.04 \times 10^{5}J $，
$ W = P_{加热}t = 2000W \times 6 \times 60s = 7.2 \times 10^{5}J $，
$ \eta = \frac{Q_{吸}}{W} = \frac{5.04 \times 10^{5}J}{7.2 \times 10^{5}J} = 70\% $。
2. (2024·潍坊)中国茶文化源远流长。图示是一款智能煮茶机的简化电路图,电源电压$U= 220V,R_{0}$、$R_{1}$、$R_{2}$是阻值不变的电热丝,用于“加热”或“保温”,$R_{1}= 20.5Ω,R_{2}= 840Ω$,S为电源开关,通过温控开关$S_{1}$可实现“加热”和“保温”状态的切换,煮茶机的“加热”功率$P_{加热}= 800W$,“加热”效率$η=80\% $。煮茶机会根据用户选择的茶品和所需茶水浓度不同进行智能工作,其工作过程是首先将水“加热”至$100^{\circ }C$,然后继续以“加热”功率再进行相应时间的“煮茶加热”(“煮茶加热”时间如表格所示),最后自动进入“保温”状态。将0.8L、初温为$20^{\circ }C$的水装入煮茶机。
(1)求将水“加热”至$100^{\circ }C$的过程中水吸收的热量;
(2)求煮茶机的“保温”功率；
(3)若用户选择“红茶高浓度煮茶”,煮茶机经过将水“加热”至$100^{\circ }C$、“煮茶加热”和“保温”状态,共工作15min,求这个过程煮茶机消耗的电能。
(1)求将水“加热”至$100^{\circ }C$的过程中水吸收的热量;
(2)求煮茶机的“保温”功率；
(3)若用户选择“红茶高浓度煮茶”,煮茶机经过将水“加热”至$100^{\circ }C$、“煮茶加热”和“保温”状态,共工作15min,求这个过程煮茶机消耗的电能。
答案
解：(1)$ m = \rho V = 1.0 \times 10^{3}kg/m^{3} \times 0.8 \times 10^{-3}m^{3} = 0.8kg $，
$ Q = cm\Delta t = 4.2 \times 10^{3}J/(kg \cdot ^{\circ}C) \times 0.8kg \times (100^{\circ}C - 20^{\circ}C) = 2.688 \times 10^{5}J $；
(2)加热时 $ R_{01} = \frac{U^{2}}{P_{加热}} = \frac{(220V)^{2}}{800W} = 60.5\Omega $，
$ R_{0} = R_{01} - R_{1} = 60.5\Omega - 20.5\Omega = 40\Omega $，
保温时 $ R_{02} = R_{0} + R_{2} = 40\Omega + 840\Omega = 880\Omega $，
$ P_{保温} = \frac{U^{2}}{R_{02}} = \frac{(220V)^{2}}{880\Omega} = 55W $；
(3)$ W_{1} = \frac{Q}{\eta} = \frac{2.688 \times 10^{5}J}{80\%} = 3.36 \times 10^{5}J $，
$ t_{1} = \frac{W_{1}}{P_{加热}} = \frac{3.36 \times 10^{5}J}{800W} = 420s = 7min $，
$ t_{2} = 5min = 300s $，
$ t_{3} = 15min - 5min - 7min = 3min = 180s $，
$ W = W_{1} + P_{加热}t_{2} + P_{保温}t_{3} = 3.36 \times 10^{5}J + 800W \times 300s + 55W \times 180s = 5.859 \times 10^{5}J $。
$ Q = cm\Delta t = 4.2 \times 10^{3}J/(kg \cdot ^{\circ}C) \times 0.8kg \times (100^{\circ}C - 20^{\circ}C) = 2.688 \times 10^{5}J $；
(2)加热时 $ R_{01} = \frac{U^{2}}{P_{加热}} = \frac{(220V)^{2}}{800W} = 60.5\Omega $，
$ R_{0} = R_{01} - R_{1} = 60.5\Omega - 20.5\Omega = 40\Omega $，
保温时 $ R_{02} = R_{0} + R_{2} = 40\Omega + 840\Omega = 880\Omega $，
$ P_{保温} = \frac{U^{2}}{R_{02}} = \frac{(220V)^{2}}{880\Omega} = 55W $；
(3)$ W_{1} = \frac{Q}{\eta} = \frac{2.688 \times 10^{5}J}{80\%} = 3.36 \times 10^{5}J $，
$ t_{1} = \frac{W_{1}}{P_{加热}} = \frac{3.36 \times 10^{5}J}{800W} = 420s = 7min $，
$ t_{2} = 5min = 300s $，
$ t_{3} = 15min - 5min - 7min = 3min = 180s $，
$ W = W_{1} + P_{加热}t_{2} + P_{保温}t_{3} = 3.36 \times 10^{5}J + 800W \times 300s + 55W \times 180s = 5.859 \times 10^{5}J $。
