(1)水吸收的热量:Q吸=cm(t2-t1)=4.2×103J/(kg?℃)×2kg×(100℃-20℃)=6.72×105J;(2)已知:R=48.4Ω,U=220V,电热丝的功率为:P= U2 R = (220V)2 48.4Ω =1000W,若不计热量损失,则W电=Q吸=6.72×105J;通电时间:t= W P = W电 P = 6.72×105J 1000W =672s.答:(1)水吸收的热量是6.72×105J(2)通电时间是672s.