(1)由P=
可知,U2 R
电热丝的电阻:R=
=U2 P
=24.2Ω;(220V)2
2000W
(2)已知VV=50L=0.05m3,
由ρ=
得:m V
m=ρV=1×103kg/m3×0.05m3=50kg,
水吸收的热量:
Q=c水m(t-t0)=4.2×103J/(kg?℃)×50kg×(45℃-25℃)=4.2×106J;
(2)热水器消耗的电能:
W=Pt=2000W×40×60s=4.8×106J,
热水器的效率:η=
×100%=Q W
×100%=87.5%.4.2×106J 4.8×106J
答:(1)热水器中电热丝的电阻为24.2Ω;
(2)热水器水箱中水所吸收的热量为4.2×106J;
(2)热水器的效率为87.5%.