略
(1)电阻丝的电阻:R=
=
=48.4Ω (2分)
(2)水吸收的热量:Q=Cm△t=4.2×103J(kg·℃)×1kg×(1
00℃-20℃)
=3.36×105J (2分)
(3)消耗的电能:W=Pt=1000W×7×60s=4.2×105J (2分)
水壶的热效率:η=
="80%" (1分)
答:略
略
(1)电阻丝的电阻:R=
=
=48.4Ω (2分)
(2)水吸收的热量:Q=Cm△t=4.2×103J(kg·℃)×1kg×(1
00℃-20℃)
=3.36×105J (2分)
(3)消耗的电能:W=Pt=1000W×7×60s=4.2×105J (2分)
水壶的热效率:η=
="80%" (1分)
答:略