2012年广西省桂林市中考物理试卷(word版)(7)

当开关S、S0闭合时，电热饮水机处于加热状态。此时R1消耗的电功率为： P1 ＝P总－ P2＝920 W－ 40W＝880 W ··························································· (1分) 则有：R1＝U2／P1＝（220V）2／880 W＝55Ω ················································ (1分) ⑶方法一：电热饮水机处于加热状态时的总电阻为：

R总＝U2／P总＝（220V）2／920W＝1210/23Ω≈ 52.6Ω ······································ (1分)

2实际加热功率：P实＝U实/R总

＝（198V）2/（1210/23）Ω＝745.2W ··································································· (1分) ［或P实＝U实2／R总＝（198V）2／52.6Ω≈745.3 W ］

方法二：电路电阻不变，可得：

R总＝U2／P总＝U实2／P实 ················································································· (1分) 实际加热功率：P实＝P额×U实2 /U2

＝920 ×（198/220）2＝745.2 W ········································································· （1分) 方法三：实际加热功率：P实＝P1实＋P2实＝U实2／R1＋U实2／R2 ···················· (1分) ＝（198V）2／55Ω＋（198V）2／1210Ω

＝745.2 W ············································································································· (1分)

28．解：1 m3可燃冰转化生成的甲烷气体完全燃烧放出热量：

Q＝q甲烷V ＝3.6×107J/ m3×164 m3＝5.904×109 J ··············································· (1分) 由题知： t＝1640min＝9.84×104s

由 P＝W / t 可得 (1分)

发动机的实际功率：P＝Q / t＝5.904×109 J／9.84×104s＝6.0×104 W ················· (2分) 由题知：36km/h＝10m/s

由 v＝s／t 可得 ··························································································· （1分) 公交车行驶路程：s＝v t＝10m/s×9.84×104s＝9.84×105 m ································· (1分) 车匀速行驶，有：

F牵＝f＝0.05 G车＝0.05mg＝0.05×6000 kg×10N/kg＝3×103N ··························· (1分) 由W＝Fs可得牵引力做的有用功： ································································· (1分) W有＝F牵s＝3×103N×9.84×105 m＝2.952×109 J ··············································· (1分) 发动机效率：

η＝W有 /Q＝2.952×109 J／5.904×109J＝50% ···················································· (1分)