(1)电热丝正常工作电阻: R=
U 2
p =
(220v) 2
2000w =24.2Ω ;
(2)水吸收的热量Q=c 水m(t-t 0)=4.2×10 3J/(kg.℃)×50kg×20℃=4.2×10 6J;
(3)热水器消耗的电能:W=Pt=2000w×2400s=4.8×10 6J;
热水器效率: η=
Q
W =
4.2× 10 6 J
4.8× 10 6 J =87.5% .
答:(1)热水器中电热丝在正常工作时的电阻值是24.2Ω;
(2)水所吸收的热量是4.2×10 6J;
(3)热水器的效率是87.5%.