(1)烧水时消耗天然气的质量:
m=ρV=0.7kg/m 3×0.01m 3=0.007kg;
(2)水吸收的热量:
Q 吸=cm(t 2-t 1)
=4.2×10 3J/(kg•℃)×1kg×(100℃-20℃)
=3.36×10 5J;
(3)由题知,Q 吸=W 电×84%,电流做功:
W 电=
Q 吸
84% =
3.36× 10 5 J
84% =4×10 5J
通电时间:
t=
W
P =
4× 10 5 J
800W =500s.
答:(1)需要消耗的煤气质量是0.007kg
(2)水吸收的热量是3.36×10 5J
(3)通电时间是500s.