第 1 页:试题 |
第 8 页:答案 |
26.解:⑴柑橘排水质量:
m排=m总-m剩=360g-240g=120 g ····································· (1分)
⑵由 ρ=m/V 可得 ····································································· (1分)
柑橘的体积:V橘 =V排=m排/ρ水=120 g/1.0 g/cm3=120 cm3 (2分)
柑橘的密度:ρ橘=m橘/V橘=114g/120 cm3=0.95g/cm3 ············ (1分)
⑶偏小 ··························································································· (1分)
27.解:⑴由 P=UI 可得 ······································································· (1分)
正常加热状态下的电流:I= P/U = 920W/220V ≈ 4.2A ··············· (1分)
⑵当开关S闭合、S0断开时,电热饮水机只有R2工作,处于保温状态。
由 P=UI I=U/R 可得 ························································ (1分)
电阻R2消耗的功率:P2=U2/R2=(220V)2/1210Ω=40W ····· (1分)
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)
实际加热功率:P实=U实2/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)
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