1、冰箱消耗的电能 W=转数n/电能表的参数N =3转/(1500转/kwh) =0 .002kwh=2×10的负3次方kwh =2×10的负3次方×3 .6×10的6次方J =7200J 2、冰箱和热水壶正常工作,一分钟电能表的转盘转33圈,冰箱单独正常工作时,一分钟电能表的转盘转3圈,则热水壶单独正常工作时电能表的转盘转30圈.热水壶消耗的电能 W=转数n/电能表的参数N =30转/(1500转/kwh) =0 .02kwh=2×10的负2次方kwh =2×10的负2次方×3 .6×10的6次方J =72000J 热水壶的电功率 P=W/t=72000J/60s =1200w 3、水吸收的热量 Q=cm(t—to) =4200J/kg℃×1kg×(100℃—20℃) =336000J 电水壶消耗的电能W=Pt 不考虑热量的损失,Q=W.通电时间t=W/P =336000J/1200w =280s